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Inessa [10]
3 years ago
6

I NEED HELP RN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!
What is the simplified form of the square root of 144x36?


12x18
72x6
72x18
Mathematics
2 answers:
Scrat [10]3 years ago
3 0
I'm pretty sure the second one bro
s344n2d4d5 [400]3 years ago
3 0
It is 72x18!
144 divided by 2 is 72, and 36 divided by 2 is 18.
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Hypothesis Testing for Means with Small Samples
Scorpion4ik [409]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: volume of root beer in a Windsor Bottling Company can.

A sample of n=24 cans was taken and their contents measured, resulting:

X[bar]= 11.4 oz

S= 0.62 oz

Assuming that the variable has a normal distribution X~N(μ;σ²), the parameter of interest is the average contents of the root beer cans of the Windsor Bottling Company (μ)

The claim is that the population mean content of the cans is different from 12 oz, symbolically: μ ≠ 12

The statistical hypothesis (Null and alternative) have to be complementary, exhaustive and mutually exclusive. The null hypothesis is the "no change" hypothesis and always carries the "=" sign.

If the claim is μ ≠ 12, its complement is μ = 12, the expression carrying the "=" sign will be the null hypothesis and its complement will be the alternative hypothesis:

H₀: μ = 12

H₁: μ ≠ 12

α: 0.05

To test the population mean of this normal population, you have to apply a one sample t-test, with statistic:

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~t_{n-1}

t_{H_0}= \frac{11.4-12}{\frac{0.62}{\sqrt{24} } } = -4.74

This test is two-tailed, using the critical value approach, you have to determine two rejection regions. Meaning, you'll reject the null hypothesis to small values of the statistic or to high values of the statistic.

t_{n-1;\alpha /2}= t_{23;0.025}= -2.069

t_{n-1;1-\alpha /2}= t_{23;0.975}= 2.069

The decision rule is:

If t_{H_0} ≤ -2.069 or if t_{H_0} ≥ 2.069, then you reject the null hypothesis.

If -2.069 < t_{H_0} < 2.069, then you do not reject the null hypothesis.

The value is less than the left critical value, the decision is to reject the null hypothesis.

Then you can say that with a 5% significance level, there is significant evidence to reject the null hypothesis, then the average amount of root beer of the Windsor Bottling Company is different from 12 oz, this means that the claim about the amount of root beer in the cans is correct.

I hope it helps!

6 0
3 years ago
If a construction work digs down to 30 feet below the surface of the ground. What is her elevation?
tresset_1 [31]

Answer:

-30 feet

Step-by-step explanation:

You are going down so it would be negative, if you were going up it would be posotive.

4 0
3 years ago
Help me plzz please please​
Snowcat [4.5K]

Answer:

132 square feet

Step-by-step explanation:

a+b/2*h

20+24/2*6=132

4 0
3 years ago
What is the ratio 16 to 40 written as a fraction in lowest terms? plz help 50 points
Luda [366]

Answer:- \frac{2}{5}

Step-by-step explanation:

The given ration is 16 :40

In fraction, the above ratio can be written as : \frac{16}{40}

To write the above fraction in lowest terms, we need to divide the numerator and the denominator by the greatest common divisor.

Since, 16=2\times8\\40=5\times8, Hence, 8 is the greatest common divisor of 16 and 40.

Now, divide the numerator and denominator by 8, we get

\frac{16}{40}=\frac{2}{5}

3 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
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