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2 years ago

PLEASE HELP. .......

Mathematics

1 answer:

Anna11 [10]2 years ago

5 0

Answer:

Step-by-step explanation:

Can fit 4 length wise, 5 width wise, and 2 depth wise. So 40 packages total.

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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

Find the Value of X and Y. Picture attached in comments.

enyata [817]

I hope this helps you

2x+7=3x-7

x=14

2x+7+12y+1=180

2.14+8+12y=180

12y=180-36

12y=144

y=12

6 0

3 years ago

Read 2 more answers

A line intersects the point (-3, -7) and

Anastaziya [24]

Answer:

y=-3x-16

Step-by-step explanation:

y-y1=m(x-x1)

y-(-7)=-3(x-(-3))

y+7=-3(x+3)

y=-3x-9-7

y=-3x-16

8 0

2 years ago

Jackie is taking a week-long football class. The first day, she gains 6 yards on her only possession of the football. On the sec

Veronika [31]

Answer:

Step-by-step explanation:

4 0

3 years ago

Point B has coordinates (3,1). The x-coordinate of point A is negative 1. The distance between point A and point B is 5 unit

mina [271]

Answer:

Given in quesion:Point B (3,1) and point A is -1. Distance between A and B is 5 units.

Using distance formula for AB we obtain,

AB^2=(x2-x1)^2 + (y2-y1)^2

5^2=(3+1)^2 + (1-y)^2

25-16=(1-y)^2

9=(1-y)^2

1-y=3

y= -2

Therefore the possible co-ordinates of A are(-1,-2).This is your required answer.

3 0

3 years ago