Multiply c^2(c^2-10c+25)

Mary catherine made 4 gallons of punch for her party how many cups of punch did she make

zhenek [66]

Answer:

64 cups

Step-by-step explanation:

1 gallon is equivalent to 16 cups

4 0

3 years ago

Read 2 more answers

Suppose that the speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles p

denis23 [38]

Answer:

A) $X \sim N(68 , 25)$

B) the probability that is traveling more than 70 mph is 0.3446

C) the probability that it is traveling between 65 and 75 mph is 0.6449

D) 90% of all cars travel at least 56.85 mph fast on the freeway

Step-by-step explanation:

The speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles per hour.

Mean = $\mu = 68 mph$

Standard deviation = $\sigma = 5 mph$

A) X ~ N( _____, _______ )

In general $X \sim N( \mu , \sigma^2)$

$\mu = 68 mph$

$\sigma = 5 mph$

$\sigma^2 = 5^2 = 25$

So, $X \sim N(68 , 25)$

B) If one car is randomly chosen, find the probability that is traveling more than 70 mph.i.e.P(X>70)

So, $Z = \frac{x-\mu}{\sigma}\\Z=\frac{70-68}{5}$

Z=0.4

Using Z table

P(Z>70)=1-P(Z<70)=1-0.6554=0.3446

Hence the probability that is traveling more than 70 mph is 0.3446

C) If one of the cars is randomly chosen, find the probability that it is traveling between 65 and 75 mph.

P(65<X<75)

$Z = \frac{x-\mu}{\sigma}$

AT x = 65

$Z=\frac{65-68}{5}$

Z=-0.6

AT x = 75

$Z=\frac{75-68}{5}$

Z=1.4

Using Z table

P(65<X<75)=P(-0.6<Z<1.4)=P(Z<1.4)-P(Z<-0.6)=0.9192-0.2743=0.6449

Hence the probability that it is traveling between 65 and 75 mph is 0.6449

D)90% of all cars travel at least how fast on the freeway?

Since we are supposed to find at least how fast on the freeway

So,P(X>x)=0.9

1-P(X<x)=0.9

1-0.9=P(X<x)

0.1=P(X<x)

Z value at 10% =-2.23

So, $Z=\frac{x-\mu}{\sigma}\\-2.23=\frac{x-68}{5}\\-2.23 \times 5 =x-68\\(-2.23 \times 5)+68=x$

56.85 = x

Hence 90% of all cars travel at least 56.85 mph fast on the freeway

8 0

3 years ago

A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le

CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁: μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

$Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2} } }$