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Dafna11 [192]
3 years ago
13

Pls i need this, it’s due today. i will mark brainliest.

Mathematics
1 answer:
MArishka [77]3 years ago
8 0

Answer:

Step-by-step explanation:

Y intercepts : 5,-1,3,7,1

X- intercepts: 2,3,1,-1

Its decreasing

It is a negative slope

Maximum : 7 for y

Min : -1

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In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

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

Compute the standard error value as follows:

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

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
Nobody is helping me on this pls answer and ty
Aleks [24]

Answer:

38.7

Step-by-step explanation:

\frac{sinA}{A}=\frac{sinB}{B}=\frac{sinC}{C}

A, B and C are the sides of the triangle and sinA, sinB and sinC are the opposing angles

\frac{sin95}{43}=\frac{A}{27}

A = sin^{-1} (\frac{27*sin95}{43} )=38.72018809

3 0
3 years ago
The equation of the line with slope = 3, going through point (2, 4) is:
Marina86 [1]

(\stackrel{x_1}{2}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{-3}(x-\stackrel{x_1}{2}) \\\\\\ y-4=-3x+6\implies y=-3x+10

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How can u solve the surface area and volume ?
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Use these formulas to get the surface area and volume. Hopes this helps! ❤️

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A rice calls for 16 fluid ounces of light whipping cream. if anthony has 1 pint of whipping cream in his refrigerator, does he h
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Yes he has enough because 16 ounces= one pint
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2 years ago
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