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2 years ago

1. A line that passes through the point (-1, 3) has a slope of 2. Find another point on the line.

Mathematics

2 answers:

Stella [2.4K]2 years ago

8 0

Answer:

option Bis correct answer.

Step-by-step explanation:

given, a line passes through the point (_1,3)and has slope 2.let another point be x2 and y2.

by the slope formula,

slope (m)=

$y2 - y1 \div x2 - x1$

or , 2=y2_3/x2_(_1)

or, equating with slope we get,

2=y2_3

or, y2=5

again, 2=x2+1

therefore, x2=1

so the points are (1,5)........ans....

dedylja [7]2 years ago

8 0

The answer is B (1,5)

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A sampling of hemlock trees revealed that 7 out of 190 trees are infected with insects. About how many of the 950 hemlock trees

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Answer:

About 35 trees in a state park are infected with insects

Step-by-step explanation:

we know that

7 out of 190 trees are infected with insects

using proportion

Find out about how many of the 950 hemlock trees in a state park are infected with insects

$\frac{190}{7}=\frac{950}{x} \\\\x=7(950)/190\\\\x= 35$

therefore

About 35 trees in a state park are infected with insects

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3 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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Going out on a limb here and guessing that the function is

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