Help!!!!!! See image.

Answer:

0.6826

Step-by-step explanation:

Probabilities are used to determine the chances of events.

The probability that at least 2 animals are infected is (c) 0.913

The proportion of the animal infected is given as:

[tex]p =60\%[/tex]

The probability is then calculated using the following binomial equation

[tex]P(x) = ^nC_xp^x(1-p)^{n-x}[/tex]

In this case,

[tex]n = 2[/tex]

To calculate the probability that at least 2 animals are infected, we start by calculating the probability that not up to 2 animals are infected.

So, we have:

[tex]P(x<2) =P(0) + P(1)[/tex]

This gives

[tex]P(x<2) = ^5C_0 \times (60\%)^0 \times (1-60\%)^{5-0}+ ^5C_1 \times (60\%)^1 \times (1-60\%)^{5-1}[/tex]

Simplify

[tex]P(x<2) = 1 \times (60\%)^0 \times (40\%)^{5}+ 5 \times (60\%) \times (40\%)^{4}[/tex]

[tex]P(x<2) = 0.08704[/tex]

Using the complement rule, we have:

[tex]P(x \ge 2) = 1 - P(x<2)[/tex]

So, we have:

[tex]P(x \ge 2) = 1 - 0.08704[/tex]

[tex]P(x \ge 2) = 0.91296[/tex]

Approximate

[tex]P(x \ge 2) = 0.913[/tex]

Hence, the probability is (c) 0.913

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Answer:

Volume is 60ft³

Step-by-step explanation:

b²h/3

b=6

h=5

The volume of a square-based pyramid will be 180 ft³. Option D is correct.

If the prism is such that if we slice it horizontally at any height smaller or equal to its original height, the cross-section is the same as its base, then its volume is:

[tex]\rm V = B^2 \times h \\\ V = 6^2 ft \times 5 \ ft \\\ V=180 \ ft^2[/tex]

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#SPJ2

Step-by-step explanation:

We have cartisean points. We are trying to find polar points.

We can find r by applying the pythagorean theorem to the x value and y values.

[tex]r {}^{2} = {x}^{2} + {y}^{2} [/tex]

And to find theta, notice how a right triangle is created if we draw the base(the x value) and the height(y value). We also just found our r( hypotenuse) so ignore that. We know the opposite side and the adjacent side originally. so we can use the tangent function.

[tex] \tan(x) = \frac{y}{x} [/tex]

Remeber since we are trying to find the angle measure, use inverse tan function

[tex] \tan {}^{ - 1} ( \frac{y}{x} ) = [/tex]

Answers For 2,5

[tex] {2}^{2} + {5}^{2} = \sqrt{29} = 5.4[/tex]

So r=sqr root of 29

[tex] \tan {}^{ - 1} ( \frac{5}{2} ) = 68[/tex]

So the answer is (sqr root of 29,68).

For -3,3

[tex] { -3 }^{2} + {3}^{2} = \sqrt{18} = 3 \sqrt{2} [/tex]

[tex] \tan {}^{ - 1} ( \frac{3}{ - 3} ) = - 45[/tex]

Use the identity

[tex] \tan(x) = \tan(x + \pi) [/tex]

So that means

[tex] \tan(x) = 135[/tex]

So our points are

(3 times sqr root of 2, 135)

For 5,-3.5

[tex] {5}^{2} + {3.5}^{2} = \sqrt{37.25} [/tex]

[tex] \tan {}^{ - 1} ( \frac{ - 3.5}{ - 5} ) = 35[/tex]

So our points are (sqr root of 37.25, 35)

For (0,-5.4)

[tex] {0}^{2} + { - 5.4}^{2} = \sqrt{} 29.16 = 5.4[/tex]

So r=5.4

[tex] \tan {}^{ - 1} (0) = undefined[/tex]

So our points are (5.4, undefined)

Answer:

There monthly income is 300$

Answer:

Hence, the height of the tree is 14.3 feet.

Answer:

"less than high school" and" high school graduate".

Answer:

The table below is comparing level of education achieved to the rate of unemployment and the median weekly earnings in 2008.  Based on the information provided, the unemployment rate decreases the most when moving between which two consecutive educational levels?

a.

“Less than high school” and “High School Graduate”  <<<<<<<CORRECT

b.

“High school graduate” and “Some college, no degree”

c.

“Associate degree” and “Bachelor’s degree”

d.

“Professional degree” and “Doctoral degree”

Step-by-step explanation:

Edge2021

Answer:

y is directly proportional to 8

while x is inversily proportional to 8

Answer:

Solution;

given equation : 3.67 = c - 2.13

On solving equation, we get

c = 3.67 +2.13

c = 5.8

Hence, Janae solution is incorrect. So I disagree.

Answer: No because the answer is c =5.8 or 5.80

Step-by-step explanation:

Add 2.13 to both sides. 3.67 + 2.13 = 5.80

I don't what form of measurement to use (like grams, cusp, etc.) but I'm pretty sure it'd be cups

I think it's 5/96 or .05208333

So, 5/8 is .625. Dividing that up in throwing it into 12 cupcakes.

9514 1404 393

Answer:

  6.2

Step-by-step explanation:

We presume your "k-value" is the k in the exponential decay term ...

  e^(-kt) . . . where t is the number of time units

This is 1/2 when ...

  ln(1/2) = -kt

  t = ln(1/2)/(-k) = ln(2)/k

  t = 0.69315/0.1124 ≈ 6.2

The half life is about 6.2 time units.

Answer:

130

Step-by-step explanation:

3*10 =30+100=130

Answer:

X = 6

Step-by-step explanation:

9514 1404 393

Answer:

  643.72 m²

Step-by-step explanation:

The area is the sum of the areas of the parts. The figure can be divided into a triangle and a semicircle.

The formula for the area of the triangle is ...

  A = 1/2bh

For the given base of 28 m and height of 24 m, the area is ...

  A = 1/2(28 m)(24 m) = 336 m²

__

The area of a semicircle is given by the formula ...

  A = (π/8)d² . . . . for diameter d

The given diameter is 28 m, so the area is about ...

  A = (3.14/8)(28 m)² = 307.72 m²

__

Then the total area of the figure is ...

   total area = triangle area + semicircle area

   total area = 336 m² +307.72 m² = 643.72 m²

Answer:

-3,4

Step-by-step explanation:

just follow up the other coordinates

Solution :

The null and the alternate hypothesis can be stated as :

Null hypothesis

[tex]$H_0:\mu_1 \geq \mu_2$[/tex]

Alternate hypothesis

[tex]$H_a:\mu_1 \leq \mu_2$[/tex]

We known;

[tex]$\overline x_1=\frac{\sum_{i=1}^n X_i}{n_1}$[/tex]

   [tex]$=\frac{43+....+75}{7}$[/tex]

   = 62.286

[tex]$\overline x_2=\frac{\sum_{i=1}^n X_i}{n_2}$[/tex]

   [tex]$=\frac{66+....+81}{16}$[/tex]

   = 72.125

[tex]$s_1^2=\frac{\sum_{i=1}^n(X_i- \overline X_1)^2}{n_1-1}$[/tex]

    [tex]$=\frac{(43-65.5)^2+....+(75-65.5)^2}{7-1}$[/tex]

   = 116.571

[tex]$s_2^2=\frac{\sum_{i=1}^n(X_i- \overline X_2)^2}{n_2-1}$[/tex]

    [tex]$=\frac{(66-72.13)^2+....+(81-72.13)^2}{16-1}$[/tex]

   = 23.45

Therefore, calculating the test statics :

[tex]$t=\frac{\overline x_1 - \overline x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$[/tex]

[tex]$t=\frac{62.29-72.125}{\sqrt{\frac{116.571}{7}+\frac{23.45}{16}}}$[/tex]

[tex]$t=\frac{-9.839}{4.2566}$[/tex]

 = -2.312

Now calculating the P-value for the test as follows :

P=T.DIST(t, df)

[tex]$df=\frac{\left(\frac{s_1^2}{n_1}+\frac{s^2_2}{n_2}\right)^2}{\frac{1}{n_1-1}\left(\frac{s^2_X}{n_1}\right)^2+\frac{1}{n_2-1}\left(\frac{s^2_Y}{n_2}\right)^2}$[/tex]

[tex]$df=\frac{\left(\frac{116.571}{7}+\frac{23.45}{16}\right)^2}{\frac{1}{7-1}\left(\frac{116.571}{7}\right)^2+\frac{1}{16-1}\left(\frac{23.45}{16}\right)^2}$[/tex]

    [tex]$=\frac{328.2868}{46.36395}$[/tex]

   [tex]$\approx 7$[/tex]

P=T.DIST(t, df)

 =T.DIST(-2.31, 7)

 = 0.0270

Thus, the [tex]$\text{P-value}$[/tex] of the test is P = 0.0270 is [tex]$\text{less}$[/tex] than the level of significance [tex]$\alpha= 0.05$[/tex]. Hence the researcher can reject the null hypothesis.

Conclusion: The mean launch temperature for the flights with O ring damages less than that for the flights with no O rings.

Answer:

B

Step-by-step explanation:

because if the car is traveling at 60 mph, the distance traveled would be 60 x the number of hours

Answer:

b

Step-by-step explanation:

I took the exam hopes this helps

Answer:

lesser x = -2, greater x = 2/3

Step-by-step explanation:

(2x+4)= 0 ---> bring 4 to the other side

2x=-4 ----> divide by 2 on both sides

x= -4/2 = -2 ----> simplify

(3x-2)= 0 ----> bring the 2 to the other side

3x=2 ----> divide by three on both sides

x=2/3

hope this helps!

Answer:

Joseph has 90 coins after 3 months

Step-by-step explanation:

Joseph: 25x + 15

Molly: 25x + 25

x = 3

Joseph: 25(3) + 15  =  90

Gravitational force = (gravity x mass1x mass2)/distance^2

6.0x10^-10 = (6.67x10^-11 x 9 x 9)/d^2

D = sqrt(6.0x10^-10/6.67x10^-11 x81)

D = 3.0 m

The answer is B.3.0 m

Answer:

B

Step-by-step explanation:

3.0 m

Answer:

m∠RPQ = 8°

Step-by-step explanation:

m∠QRS = 4x - 15

m∠RPQ = x + 1

m∠PQR = x - 2

m∠QRS is exterior angle and m∠RPQ and m∠PQr are opposite interior angles to m∠QRS

m∠QRS = m∠RPQ + m∠PQR      {Exterior angle property of triangle}

4x - 15 = x +1 + x - 2

4x - 15 = x + x + 1-2          {Combine like terms}

4x - 15 = 2x - 1        {Subtract 2x from both sides}

4x - 2x - 15 = - 1

2x - 15 = - 1                  {Add 15 to both sides}

2x = -1 + 15

2x = 14          {Divide both sides by 2}

x = 14/2

x = 7

m∠RPQ = x + 1 = 7 + 1 = 8°

Answer: x+1 =(7)+1=8

Step-by-step explanation:

Answer:

Echo stop DELeting MY Stuff

Step-by-step explanation:

Answer:

option 2.

the others are simultaneous equations in two unknowns but 7x-2y have the same value and the value can't be two different things at once

30%....................................

Answer:  At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68

Step-by-step explanation: u divide

Answer:

C) 28

Step-by-step explanation:

RSTV = WXYZ

WX = x

[tex]\frac{RS}{VR} =\frac{WX}{ZW} \\\\\frac{7}{3} =\frac{x}{12} \\\\3x=84\\x=28[/tex]

Answer:

-5+5i

Step-by-step explanation:

Answer:

D on edge

Step-by-step explanation:

-5+5i

Answer:

132

Step-by-step explanation: I'm not very sure tho

9514 1404 393

Answer:

  (a) -6±√5

Step-by-step explanation:

  (z +6)^2 = 5 . . . . . given

  z +6 = ±√5 . . . . .  take the square root

  z = -6 ±√5 . . . . . . subtract 6

Answer:

A parallelogram has opposite sides parallel and equal in length. Also opposite angles are equal (angles "A" are the same, and angles "B" are the same). NOTE: Squares, Rectangles and Rhombuses are all Parallelograms!

Yes, they are.

Answer:

Yes

Step-by-step explanation:

Squares are quadrilaterals with 4 congruent sides and 4 right angles, and they also have two sets of parallel sides. Parallelograms are quadrilaterals with two sets of parallel sides. Since squares must be quadrilaterals with two sets of parallel sides, then all squares are parallelograms.

• ||v|| = 42, which is to say

||v||² = 〈v, v 〉

…  = 〈a u₁ + b u₂ + c u₃, a u₁ + b u₂ + c u₃〉

… = a ² 〈u₁, u₁〉 + b ² 〈u₂, u₂〉 + c ² 〈u₃, u₃〉 + 2(ab 〈u₁, u₂〉 + ac 〈u₁, u₃〉 + bc 〈u₂, u₃〉)

… = a ² ||u₁||² + b ² ||u₂||² + c ² ||u₃||²

[since each vector in the basis for V is orthogonal to any other vector in the basis, and 〈x, x〉 = ||x||² for any vector x ]

42² = a ² + b ² + c ²

[since each vector in the basis has unit length]

42 = √(a ² + b ² + c ²)

• v is orthogonal to u₃, so 〈v, u₃〉 = 0. Expanding v gives the relation

〈v, u₃〉 = 〈a u₁ + b u₂ + c u₃, u₃〉

… = a 〈u₁, u₃〉 + b 〈u₂, u₃〉 + c 〈u₃, u₃〉

… = c ||u₃||²

… = c

which gives c = 0, and so

42 = √(a ² + b ²)

• Lastly, 〈v, u₂〉 = -42, which means

〈v, u₂〉 = 〈a u₁ + b u₂ + c u₃, u₂〉

… = a 〈u₁, u₂〉 + b 〈u₂, u₂〉 + c 〈u₃, u₂〉

… = b ||u₂||²

… = b

so that b = -42. Then

42 = √(a ² + (-42)²)   →   a = 0

So we have a = 0, b = -42, and c = 0.

The required values are, [tex]a=0,b=-42 ,c=0[/tex]

Given,

[tex]v=au_1+bu_2+cu_3[/tex]

[tex]\left\| V\right\|=42[/tex]

Computation:

Since, [tex]v[/tex] is orthogonal to [tex]u_3[/tex] then we have,

[tex]\left<v,u_3 \right> =0\\\left< v,u_2\right> =-42[/tex]

Then,

[tex]\left\| V\right\|^2=\left<v,v \right>\\=\left<au_1+bu_2+cu_3,au_1+bu_2+cu_3 \right>\\=a_2\left\|u_1 \right\|^2+b_2\left\|u_2 \right\|^2+c_2\left\|u_3 \right\|^2\\=a^2+b^2+c^2\\=a^2+b^2+c^2=42^2[/tex]

As we know,

[tex]a=\left<v_1u_1 \right>\\b=\left< v_1u_2\right>= -42\\c=\left<v_1u_3 \right> =0[/tex]

[tex]a_2+b_2+c_2=42\\a=0[/tex]

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