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3 years ago

8

A random sample of size n1 = 16 is selected from a normal population with a mean of 75 and standard deviation of 8. A second ran

dom sample of size n2 = 9 is taken independently from another normal population with mean 70 and standard deviation of 12. Let X1 and X2 be the two sample means. Find
(a) The probability that X1 − X2 exceeds 4.
(b) The probability that 3.5 < X1 − X2 < 5.5.

1 answer:

gavmur [86]3 years ago

4 0

Answer:

the answer is in the attached image below

Step-by-step explanation:

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When Paula went bowling, she scored 118 and 138 in her

Lena [83]

Answer:

140

Step-by-step explanation:

yes

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3 years ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

3 years ago

Scientific notation 100400

ad-work [718]

1.004 * 10^5
all scientific notation equations have to be a number greater than 1 but less than 10
then multiplied by 10 to the power of (move the decimal point to the right however many times you need, in this case 5) :)

5 0

3 years ago

Use the given values of n and p to find the minimum usual value μ−2σ and the maximum usual value μ+2σ. Round to the nearest hund

lbvjy [14]

Answer:

μ−2σ = 1,089.26

μ+2σ = 1,097.62

Step-by-step explanation:

The standard deviation of a sample of size 'n' and proportion 'p' is:

$\sigma=\sqrt{\frac{p*(1-p)}{n} }$

If n=1139 and p =0.96, the standard deviation is:

$\sigma=\sqrt{\frac{p*(1-p)}{n}}\\\sigma = 0.001836$

The minimum and maximum usual values are:

$\mu-2\sigma = (p-2\sigma)*n\\\mu+2\sigma = (p+2\sigma)*n$

$\mu-2\sigma = (0.96-2*0.001836)*1139\\\mu-2\sigma = 1,089.26\\\mu+2\sigma = (0.96+2*0.001836)*1139\\\mu+2\sigma = 1,097.62$

5 0

3 years ago

Malcolm bought 6 bowls for $13.20. whats the unit rate<br><br><br> need the work

Whitepunk [10]

Hello!

You have to find the cost of one bowl

To find this you take the cost of all the bowls and divide it by the amount of bowls

13.20 / 6 = 2.2

The answer is $2.20

Hope this helps!

8 0

3 years ago

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