How much bigger is 9 3/4 than 9 5/8

Answer:

the last one- 4/24

Step-by-step explanation:

ive answered this question before and got it correct

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Answer:

50%

Step-by-step explanation:

Suppose you have a fair coin; this means it has a 50% chance of landing heads up. Suppose you dip it three times, these flips are independent. What is the probability that it lands heads up, then tails up and then heads up again. so the answer is 1/8 or 12.5%

(hope I wasnt late!!)

Answer:

If a = 5 and b = 3, then that means 2a - 3b + 3a = 16

Step-by-step explanation:

2a - 3b + 3a

(2×5) - (3×3) + (3×5)

10 - 9 + 15 = 16

Answer:

sorry, I think u got yr question incomplete.

To find the median:

- Arrange the data points from smallest to largest.

- If the number of data points is odd, the median is the middle data point in the list.

- If the number of data points is even, the median is the average of the two middle data points in the list.

Answer:

POINTS HA

Step-by-step explanation:

Answer:

125 inches cubed

Step-by-step explanation:

First, you find the volume of the full square, and then you divide your answer by 2, since the figure is a triangle.

So,

10 x 10 x 2.5 = 250

250 ÷ 2 = 125 inches cubed

Answer:

the ans is 21 hope it may help u

Step-by-step explanation:

g = 6

x=6

x= 3x + 3

x = 3 × 6 +3

x = 18 + 3

x= 21

Answer:

6 1/2 quarts I think

Step-by-step explanation:

So sorry if I'm wrong! Have a nice day! Owa Owa!

Answer:

In (16.8+12=) 28.8 minutes the person will be left only with 2mg of dye in the body and be able to leave the doctor's office after being injected with 11 mg

Step-by-step explanation:

After 12 minutes, 7.75 milligrams of dye remain in your system.

This means that 11- 7.75= 3.25 milligrams of dye are used up in 12 minutes

7.75-2= 5.75 milligrams still needs to be used up

         Dye           Minutes

         3.25             12

          5.75             x

x= 12*5.75/3.25

x= 16.8 minutes

5.75 mg will be used up in 16.8 minutes

In (16.8+12=) 28.8 minutes the person will be left only with 2mg of dye in the body and be able to leave the doctor's office.

Answer:

hope it helps ya.

please give me brainliest

answer= 1

5 x 2 - 3^2

1. First simplify exponents

= that is nine

2. Then Do 5 x 2- 9

5 x 2=10

10-9

= 1

What's your numerator and denominator value?

Answer: 720 ways

Step-by-step explanation:

Given

Leroy wants to teach his puppy 6 new tricks

Considering each trick is different from other

The first trick can be taught in 6 different ways

After learning the first trick, there are 5 tricks remaining which can be taught in 5 different ways

Similarly, for the remaining tricks, it is 4, 3, 2, and 1 way

So, the total number of ways is [tex]6\times 5\times 4\times 3\times 2\times 1=720\ \text{Ways}[/tex]

Answer:

B 18.38

Step-by-step explanation:

log4x = 2.1

logx/log4 = 2.1

logx = log(4) x 2.1

logx = 1.2643

x = antilog(1.2643)

x = 18.38 (to 2 d.p)

Answer:

X=18.38

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

1. x=16

2. x=35

3. x=10

4. x=30

5. x=3

Step-by-step explanation:

intercepted arcs are 2* the angle measure

1. x= 1/2( 32)=16 x=16

2. the intercepted arc is 360-110-180=70

1/2(70)= 35

3. 2x=x+10  subtract x from both sides

x=10

4. 2(x+30)=4x

2x+60=4x

60=2x

30=x

5. 2(3x)=12x-18

6x=12x-18

-6x=-18

x=m

Answer:

Can you add a picture if there are any?

Step-by-step explanation:

Answer:

the 2nd and 3rd one I believe

Answer:

Step-by-step explanation:

Answer:

0.3425 = 34.25% probability it will be off probation in February 2020

Step-by-step explanation:

We have these desired outcomes:

Off probation in July 2019, with 0.25 probability, then continuing off in January, with 1 - 0.08 = 0.92 probability.

Still in probation in July 2019, with 1 - 0.25 = 0.75 probability, then coming off in January, with 0.15 probability.

What is the probability it will be off probation in February 2020?

[tex]p = 0.25*0.92 + 0.75*0.15 = 0.3425[/tex]

0.3425 = 34.25% probability it will be off probation in February 2020

Answer:

b.)36

Step-by-step explanation:

Formula for area of a trapezoid = [tex]\frac{a+b}{2} h[/tex]

where a and b = bases and h = height

The sign has the following dimensions

Base 1 = 6ft

base 2 = 12ft

height = 4ft

Using these dimensions we plug in the values into the formula

[tex]A=\frac{6+12}{2} 4\\6+12=18\\\frac{18}{2} =9\\9*4=36[/tex]

Hence the area of Mr. Wash's sign is 36 square feet.

Answer:

b) 36 sq ft.

Step-by-step explanation:

6 x 4 = 24 (multiply to find the square after splitting it)

12 - 6 = 6 x 4 = 24 / 2 = 12 (you divide by two because it's a triangle)

24 + 12 = 36

hope this helps :)

Answer:

a) 0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b) 0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

Step-by-step explanation:

For each component, there are only two possible outcomes. Either they last more than 1,000 hours, or they do not. Components operate independently, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

One subsystem has eight identical components, each with a probability of 0.1 of failing in less than 1,000 hours.

So 1 - 0.1 = 0.9 probability of working for more, which means that [tex]p = 0.9[/tex]

a. exactly two of the four components last longer than 1000 hours.

This is P(X = 2) when [tex]n = 4[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.9)^{2}.(0.1)^{2} = 0.0486[/tex]

0.0486 = 4.86% probability that exactly two of the four components last longer than 1000 hours.

b. the subsystem operates longer than 1000 hours.

The subsystem has 8 components, which means that [tex]n = 8[/tex]

It will operate if at least 4 components are working correctly, so we want:

[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]

In which

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{8,0}.(0.9)^{0}.(0.1)^{8} \approx 0[/tex]

[tex]P(X = 1) = C_{8,1}.(0.9)^{1}.(0.1)^{7} \approx 0tex]

[tex]P(X = 2) = C_{8,2}.(0.9)^{2}.(0.1)^{6} \approx 0[/tex]

[tex]P(X = 3) = C_{8,3}.(0.9)^{3}.(0.1)^{5} = 0.0004[/tex]

Then

[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0 + 0 + 0.0004 = 0.0004[/tex]

[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.0004 = 0.9996[/tex]

0.9996 = 99.96% probability that the subsystem operates longer than 1000 hours.

Answer:

100

Step-by-step explanation:

it gives you the formula for simple interest

Answer:

B

Step-by-step explanation:

Angle 1 is an exterior angle, while angle 6 is an interior angle and so they are not congruent

Answer:

[tex]E(Y)=\frac{1}{25}[/tex]

Step-by-step explanation:

Let's start defining the random variables for this exercise :

[tex]X_{1}:[/tex] '' The proportion of the particulates that are removed by the first pass ''

[tex]X_{2}:[/tex] '' The proportion of what remains after the first pass that is removed by the second pass ''

[tex]Y:[/tex] '' The proportion of the original particulates that remain in the sample after two passes ''

We know the relation between the random variables :

[tex]Y=(1-X_{1})(1-X_{2})[/tex]

We also assume that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] are independent random variables with common pdf.

The probability density function for both variables is [tex]f(x)=4x^{3}[/tex] for [tex]0<x<1[/tex] and [tex]f(x)=0[/tex] otherwise.

The first step to solve this exercise is to find the expected value for [tex]X_{1}[/tex] and [tex]X_{2}[/tex].

Because the variables have the same pdf we write :

[tex]E(X_{1})= E(X_{2})=E(X)[/tex]

Using the pdf to calculate the expected value we write :

[tex]E(X)=\int\limits^a_b {xf(x)} \, dx[/tex]

Where [tex]a=[/tex] ∞ and [tex]b=[/tex] - ∞ (because we integrate in the whole range of the random variable). In this case, we will integrate between [tex]0[/tex] and [tex]1[/tex] ⇒

Using the pdf we calculate the expected value :

[tex]E(X)=\int\limits^1_0 {x4x^{3}} \, dx=\int\limits^1_0 {4x^{4}} \, dx=\frac{4}{5}[/tex]

⇒ [tex]E(X)=E(X_{1})=E(X_{2})=\frac{4}{5}[/tex]

Now we need to use some expected value properties in the expression of [tex]Y[/tex] ⇒

[tex]Y=(1-X_{1})(1-X_{2})[/tex] ⇒

[tex]Y=1-X_{2}-X_{1}+X_{1}X_{2}[/tex]

Applying the expected value properties (linearity and expected value of a constant) ⇒

[tex]E(Y)=E(1)-E(X_{2})-E(X_{1})+E(X_{1}X_{2})[/tex]

Using that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] have the same expected value [tex]E(X)[/tex] and given that [tex]X_{1}[/tex] and [tex]X_{2}[/tex] are independent random variables we can write [tex]E(X_{1}X_{2})=E(X_{1})E(X_{2})[/tex]   ⇒

[tex]E(Y)=E(1)-E(X)-E(X)+E(X_{1})E(X_{2})[/tex] ⇒

[tex]E(Y)=E(1)-2E(X)+[E(X)]^{2}[/tex]

Using the value of [tex]E(X)[/tex] calculated :

[tex]E(Y)=1-2(\frac{4}{5})+(\frac{4}{5})^{2}=\frac{1}{25}[/tex]

[tex]E(Y)=\frac{1}{25}[/tex]

We find that the expected value of the variable [tex]Y[/tex] is [tex]E(Y)=\frac{1}{25}[/tex]

Answer:

Step-by-step explanation:

x^1/3( x^1/2 + 2x^2)

x^(1/2 + 1/3) + 2x^(2 + 1/3)

x^(5/6) + 2x^(7/3)