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Lyrx [107]
3 years ago
7

Solve for c C^2-25-11

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0
C3-36

the 3 is “c3” is an exponent btw!
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Which of the following questions is a statistical question?
stellarik [79]

Answer:

Where is the question

Step-by-step explanation: you showed no question

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3 years ago
Can someone help me please
bazaltina [42]

A: 8,250

B: 22

Step-by-step explanation:

A: 15 * 250 = 3,750 + 4,500 = 8,250

B: 10,000 - 4,500 = 5,500/250 = 22. 22 * 250 + 4,500 = 10,000

5 0
3 years ago
What is the value of tan(60°)? One-half StartRoot 3 EndRoot StartFraction StartRoot 3 EndRoot Over 2 EndFraction StartFraction 1
Ratling [72]

Answer:

\sqrt3

Step-by-step explanation:

To find:

The value of tan60^\circ = ?

Solution:

Kindly consider the equilateral \triangle ABC as attached in the answer area.

Let the side of triangle = a units

Let us draw the perpendicular from vertex A to side BC.

It will divide the side BC in two equal parts.

i.e. BD = DC = \frac{a}{2}

Using Pythagorean Theorem in \triangle ABD:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}

Side AD = \frac{\sqrt3}{2}a

Using Trigonometric ratio:

tan\theta = \dfrac{Perpendicular}{Base}

tanB = \dfrac{AD}{BD}

Putting the values of AD and BD:

tan60^\circ=\dfrac{\frac{\sqrt3}{2}a}{\frac{1}{2}a}\\\Rightarrow tan60^\circ = \bold{\sqrt3}

5 0
3 years ago
Read 2 more answers
According to Nielsen Media Research. of all the U.S. households that owned at least one television set, 83% had two or more sets
Rudik [331]

Answer:

The proportion of U.S. households that owned two or more televisions is 83%.

Step-by-step explanation:

To determine whether the proportions of U.S. households that owned two or more televisions is less than 83% or not let us perform a hypothesis test for single proportion.

<u>Assumptions:</u>

The sample size (<em>n</em>) selected by the local cable company is 300 which is quite large. Then according to the Central limit theorem the sampling distribution of sample proportion follows a normal distribution with mean <em>p</em> and standard deviation \sqrt{\frac{p(1-p)}{n} } .

Since the sampling distribution of sample proportions follows a normal distribution use the <em>z</em>-test for one proportion to perform the test.

<u>The hypothesis is:</u>

H_{0} : The proportion of U.S. households that owned two or more televisions is 83%, i.e. p=0.83

H_{1} :The proportion of U.S. households that owned two or more televisions is less than 83%, i.e. p< 0.83

<u>Decision Rule:</u>

At the level of significance α = 0.05 the critical region for a one-tailed <em>z</em>-test is:

\\ Z\leq -1.645\\

**Use the <em>z</em> table for the critical values.

So, if \\ Z\leq -1.645\\ the null hypothesis will be rejected.

<u>Test statistic value:</u>

z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}

Here \hat{p} is the sample proportion.

Compute the value of \hat{p} as follows:

\hat{p}=\frac{X}{n} \\=\frac{240}{300}\\ =0.80

Now compute the value of the test statistic as follows:

z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\=\frac{0.80-0.83}{\sqrt{\frac{0.83*(1-0.83)}{300} } } \\=-1.383

The test statistic is -1.383 which is more than -1.645.

Thus, the test statistic lies in the acceptance region.

Hence we fail to reject the null hypothesis.

<u>Conclusion:</u>

At 0.05 level of significance we fail to reject the null hypothesis stating that the proportion of U.S. households that owned two or more televisions is  83%.

6 0
3 years ago
12.
slega [8]

in a week 7

in W weeks ....W*7  this is the answerwer:

Step-by-step explanation:

7 0
4 years ago
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