X=180-2(60)
=180-120
=40
The length of the other angle is 40°.
=180-120
=40
The length of the other angle is 40°.
X^{2} +(3a+4b)x+(2a^{2} +5ab+3b^{2} )
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
Shown is one-fourth of a circle. Also, according to the picture, the radius is 28 meters.
The area for a whole circle is .
Using that information, we can find the area as if the circle was whole instead of just the fourth.
Now that we know the area of the whole circle, we just have to divide it by four to find the area of the fourth of the circle.
That means that the area of the shape shown is .
I hope this helps!
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.
