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2 years ago

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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3 years ago

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6) Work out 4 x 10-5 x 6 x 1012 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.

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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 33 degrees

The angle is given as:

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

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

$\mathbf{B = 90 - 57}$

$\mathbf{B = 33}$

Hence, the measure of acute angle is 33 degrees

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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 p = proportion of young people who had earned a high school diploma.

A sample of n = 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 - α)% confidence interval for population proportion is:

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

Compute the critical value of z 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:

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

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

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3 years ago