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AnnyKZ [126]
3 years ago
7

7. A diver dove to a location 4

Mathematics
1 answer:
dangina [55]3 years ago
7 0

Answer 1 3/5 meter below sea level

In this case, a diver goes into two different location expressed with "meters below sea level" unit. To know the distance between the two location, then you need to subtract it.

Step-by-step explanation:

Distance = 2nd location - 1st location

Distance = 8 1/5 meter below sea level - 6 3/5 meter below sea level= 1 3/5 meter below sea level

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AleksAgata [21]

Here we are given the expression:

x^{2}+18

Now let us equate it to zero to find x first,

x^{2}+18=0

Now subtracting 18 from the other side,

x^{2}=-18

taking square root on both sides,

So we will get two values of x as ,

x=3\sqrt{-2}

x=-3\sqrt{-2}

Now we can write square root -1 as i,

So our factors become,

x=3i\sqrt{2}

x=-3i\sqrt{2}

Answer:

The final factored form becomes,

(x+3i\sqrt{2})(x-3i\sqrt{2})



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I’m confused. Can you help
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Which is the best estimate for the quotient of 198.32 ÷ 9.001?
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The equation of the line of best fit of a scatter plot is y = 6x − 9. What is the slope of the equation? –6 –9 9 6
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Using traditional methods it takes 92 hours to receive an advanced flying license. A new training technique using Computer Aided
DanielleElmas [232]

Answer:

The value of the test statistic is z = 1.39.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

Step-by-step explanation:

Using traditional methods it takes 92 hours to receive an advanced flying license.

This means that at the null hypothesis, it is tested if the mean is of 92, that is:

H_0: \mu = 92

Test if there is evidence that the technique lengthens the training time

At the alternative hypothesis, it is tested if the mean is more than 92, that is:

H_1: \mu > 92

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.

92 is tested at the null hypothesis:

This means that \mu = 92

A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.

This means that n = 70, X = 93, \sigma = \sqrt{36} = 6

Value of the test statistic:

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

z = \frac{93 - 92}{\frac{6}{\sqrt{70}}}

z = 1.39

The value of the test statistic is z = 1.39.

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.

Looking at the z-table, z = 1.39 has a p-value of 0.9177.

1 - 0.9177 = 0.0823.

The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.

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