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Soloha48 [4]
2 years ago
15

X = 3y + 10 X= -2y + 15

Mathematics
1 answer:
saw5 [17]2 years ago
4 0

Answer:

Step-by-step explanation:

3y + 10 = -2y + 15

5y + 10 = 15

5y = 5

y = 1

x = 3(1) + 10

x = 3 + 10

x = 13

(13, 1)

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6) Work out 4 x 10-5 x 6 x 1012<br>Give your answer in standard form.​
julia-pushkina [17]

Answer:

4 × 5 × 6 × 1012 = 121440

= 1.2 × 10⁵

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3 years ago
Can somebody help me with this it a little hard
raketka [301]

24 3-pound bags(72 pounds) and 10 5-pound bags(50 pounds). 72+50 = 122 pounds, 24+10 bags = 34 bags.

7 0
3 years ago
Refer to the figure. Barton Road and Olive Tree Lane
Fantom [35]

Angles at right-angle add up to 90 degrees

The measure of acute angle Tryon Street forms with Barton Road is <em>33 degrees</em>

The angle is given as:

\mathbf{A = 57^o} ---- <em>Angle formed by Tryon Street with Olive Tree Lane. </em>

The measure of the acute angle formed by Tryon Street with Barton Road (B) is calculated using:

\mathbf{A + B = 90} ---- angle at right-angle

Make B the subject

\mathbf{B = 90 - A}

Substitute 57 for A

\mathbf{B = 90 - 57}

\mathbf{B = 33}

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brainly.com/question/10334248

3 0
2 years ago
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
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