hellohello, this needs to be turned in tomorrow so if you could help, i would really appreciate it. :]

Lesechka [4]

Step-by-step explanation:

Hi there!!! 

The answer is option A. 

reason look in picture. 

if you want to solve it just cross multiply it,

 $60x = 240000 \times 100$ 

or,

 $x = 24000000 \div 60$ 

and the answer would be 400000.

Hope it helps...

5 0

3 years ago

Check whether p=-2 is a solution of the equation 7p-5=20 or not ...

Lana71 [14]

Answer:

not a solution

Step-by-step explanation:

Substitute p = - 2 into the left side of the equation and if equal to the right side then it is a solution.

7(- 2) - 5 = - 14 - 5 = - 19 ≠ 20

Then p = - 2 is not a solution of the equation

8 0

2 years ago

Read 2 more answers

The Doe family is ready to fill their new swimming pool. It can be filled in 12 hours if they use their own water hose, and in 3

aivan3 [116]

6 & 1/2 hours should be the correct answer! :]

3 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!!!!

Naddik [55]

Answer:

Step-by-step explanation:

a) To find the length of diagonal XT, we can use the distance formula to get $\sqrt{(-3-5)^{2}+(4-(-2))^{2}}=10$

Since this is a rectangle, XT=YW, meaning YW=10 as well.

b) The area of a rectangle is given by length times width. The length is |-3-5|=8, and the width is |-2-4|=6. So the area is (8)(6)=48.

c) The perimeter is just 2(8+6)=28.

8 0

1 year ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

2 years ago