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lawyer [7]
3 years ago
12

Standardized​ exam's scores are normally distributed. In a recent​ year, the mean test score was 21.2 and the standard deviation

was 5.7 The test scores of four students selected at random are 15,23,7 and 35 . Find the​ z-scores that correspond to each value and determine whether any of the values are unusual.
The​ z-score for 15 is
Mathematics
1 answer:
alina1380 [7]3 years ago
8 0
32’s 21 savages lil raw jhit only
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A bridge connects to a tunnel as shown in the figure. The bridge is 180 feet above the ground, At a distance of 235 feet atong t
boyakko [2]
180^2 + 235^2 = (cliff base to point on bridge)^2
32400 + 55225 = 87625
so cliff base to point on bridge = sqrt of 87625.
Then notice 87625 + d^2 = (x+180)^2
since x^2 +235^2 = d^2,
substitute (x^2 + 235^2) for d^2, like this:
87625 + x^2 + 235^2 = (x+180)^2
87625 + x^2 + 55225 = x^2 + 360x + 32400
87625 + 55225 - 32400 = x^2 - x^2 +360x
360x = 110450
x = about 306.8 feet
h = x+180, so h = 486.8 feet
Remember d^2 = x^2 + 235^2, so
d^2 = 306.8^2 + 235^2
d^2 = 149354.6489
d = about 386.46 feet
6 0
3 years ago
For what value of n is 20 = {n+4?<br> a)<br> b) *<br> c) 40<br> d) 60
Advocard [28]
I think it might be 40 I’m not sure but good luck
6 0
3 years ago
Read 2 more answers
Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the cu
Diano4ka-milaya [45]

Answer:

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

H_0: \mu \leq 3

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

H_1: \mu > 3

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

3 is tested at the null hypothesis:

This means that \mu = 3

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

This means that n = 100, X = 3.1, \sigma = 0.5

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

z = 2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.

6 0
3 years ago
Please help ASAP!!!!!!!!!!!!
Rom4ik [11]
V = sqrt 64h

H = 76

V = sqrt * 64(76)


Answer:  V = 69.74 ft/sec






Hope that helps!!!!
4 0
3 years ago
Solve the equation. identify any extraneous solutions. x=√6x+7
marshall27 [118]
In order to solve the equation we first need to put the condition:
6x+7>=0
x>= -7/6

We square the equation:
x^2=6x+7
x^2-6x-7=0
x^2+x-7x-7=0
x(x+1)-7(x+1)=0
(x-7)(x+1)=0
x1=7
x2=1

Both solutions meet the condition.

3 0
3 years ago
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