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Lelechka [254]
3 years ago
11

HELP WILL MARK YOU BRAINLIEST HURRY IM ON A TIME LIMIT!’

Mathematics
2 answers:
Keith_Richards [23]3 years ago
6 0

(3x+5)+(x+3)=180

4x+8=180

4x=172

x=43

3(43)+5

<PQR=134

<SQR=46

levacccp [35]3 years ago
6 0

Answer:

\boxed {\angle PQR = 134\textdegree}

Step-by-step explanation:

Both angles are <u>Supplementary</u> (Two angles that add up to 180°). So, write an expression using the following measurements of both \angle PQR and \angle RQS and 180:

\angle PQR = 3x + 5

\angle RQS = x + 3

3x + 5 + x + 3 = 180

Solve:

3x + 5 + x + 3 = 180

4x + 5 + 3 = 180

4x + 8 = 180

4x + 8 - 8 = 180 - 8

4x = 172

\frac{4x}{4} = \frac{172}{4}

x = 43

-After you have the value of x, use it to the expression of \angle {PQR} to get the actual answer of \angle PQR:

-The value of x:

x = 43

-Solve for the measurement of \angle PQR:

3x + 5

3(43) + 5

129 + 5

134

-The measurement of \angle PQR:

\boxed {\angle PQR = 134\textdegree}

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A Wooden board is leaning against the house the base of the board is 10 feet from the base of the house and the base of the boar
astra-53 [7]

Answer: 12.20 feet.

Step-by-step explanation:

Observe in the figure attached that a right triangle is formed.

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cos\alpha=\frac{adjacent}{hypotenuse}

In this case you can identify that:

adjacent=10\\hypotenuse=x

\alpha=35\°

Then, to find the length of the wooden board (x), you need to substitute values and solve for x.

Therefore, you get:

cos(35\°)=\frac{10}{x}\\\\(x)(cos(35\°))=10\\\\x=\frac{10}{cos(35\°)}\\\\x=12.20

The length of the wooden board is: 12.20 feet.

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2 years ago
Is each line parallel, perpendicular, or neither parallel nor perpendicular to a line whose slope is -6?
babymother [125]

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Line M - neither

Line N - parallel

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Line Q - neither

Step-by-step explanation:

If a line is perpendicular to another, the slope will be the opposite, e.g. -6, opposite slope, -6.

If a line is parallel to another, the slope will be the exact same, e.g. -6, same slope, 1/6.

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Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
vladimir1956 [14]

Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.

The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₂₇ > 2.12) = 0.022.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0
3 years ago
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