HELP ASAP PLEASE I HAVE TO TURN THIS IN SOON

George tallies up the transactions in his check register and comes up with a total balance of $221.57, but his bank statement sa

Aneli [31]

Answer:A is the correct answer

Step-by-step explanation:

6 0

3 years ago

Simplify the expression: w2 + 7w2 + 8w - 5 + 9 - 2w2

ololo11 [35]

<h3> $6w^2 +8w +4$ is the simplified expression</h3>

<em><u>Solution:</u></em>

Given that,

We have to simplify

$w^2 + 7w^2 + 8w -5+9-2w^2$

We can simplify the above expression by combining the like terms

Like terms are terms that has same variable with same exponent and same or different coefficient

From given,

$w^2 + 7w^2 + 8w -5+9-2w^2$

Group the like terms

$w^2 + 7w^2 -2w^2 + 8w - 5+9\\\\Combine\ the\ like\ terms\\\\6w^2 +8w -5+9\\\\Combine\ the\ constants\\\\6w^2 +8w +4$

Thus the given expression is simplified

4 0

3 years ago

Luke has a calculator that will only display number less than or equal to 999,999,999. Which of the following products will his

liq [111]

Answer:

≥ 999,999,999

Step-by-step explanation:

that is the less than or equal to sign

hope this helps

8 0

2 years ago

I WILL MARK BRAINLIEST FOR THIS!!!!!!!!!!!

igomit [66]

Answer:

She gets a job as a lawyer after graduation.

Step-by-step explanation:

6 0

2 years 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

3 years ago