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kondor19780726 [428]

3 years ago

11

Urgent!!!! The scores of a high school entrance exam are approximately normally distributed with a given mean Mu = 82.4 and stan

dard deviation Sigma = 3.3. What percentage of the scores are between 75.8 and 89?
68%

95%

99.7%

100%

2 answers:

Effectus [21]3 years ago

5 0

Answer:95%

Step-by-step explanation:

DedPeter [7]3 years ago

5 0

Answer:

95 or B

Step-by-step explanation:

right on edge

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Compute 41, 42, 43, 44, 45, 46, 47, and 48. Make a conjecture about the units digit of 4n where n is a positive integer. Use str

padilas [110]

Answer:

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

Step-by-step explanation:

$4^1 = 4\\4^2 = 16\\4^3 = 64\\4^4 = 256 \\4^5 = 1024\\4^6 = 4096\\4^7 = 16384\\4^8 = 65536$

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

To prove this conjecture

$4^1 = 4\\4^2 = 16$ unit digit = 6

hence the property is true for ; n = 1 and n = 2 and also for every odd and even number ( i.e. from 1 to 8 )

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3 years ago

How many more points do I need for a gpa of a 2.0<br>

madreJ [45]

Answer:

you need at least a score of 60

Step-by-step explanation:

anything 59 and below is 1.95 and every percent higher it adds .5 more

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2 years ago

Read 2 more answers

Multiply. use the gcf. 3/4 7/9

oee [108]

21/36.

All you have to do is multiply numerators by numerators and denominators by denominators. Simplify if needed

Please consider marking branliest if this helped

7 0

3 years ago

The perimeter of a rectangle is 60 m and its length is twice its width. find the length of the rectangle

GREYUIT [131]

Length = 2W

P= 2L + 2W
60 = 2(2W) + 2W
60= 4W +2W
60 = 6W
60/6W = 6W/6W
w = 10
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Length = 20

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4 0

3 years ago

Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv

Georgia [21]

Answer:

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation

$\bar X_{1}=5.2$ represent the mean for group men

$\bar X_{2}=4.5$ represent the mean for group women

Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

$n_{1}=32$ sample size for the group men

$n_{2}=30$ sample size for the group women

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0: $\mu_{1} \leq \mu_{2}$

H1: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

$z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02$

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

5) Statistical decision

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0

3 years ago