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Nina [5.8K]
3 years ago
12

Line A is perpendicular to Line B. If the slope of A is -1/7, what is the slope of Line B?

Mathematics
1 answer:
Arlecino [84]3 years ago
7 0

Answer:

m = 7

Step-by-step explanation:

Perpendicular slopes are negative reciprocals so:

m of A = -1/7

m of B = 7

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Paul has $30,000 to invest. His intent is to earn 15% interest on his investment. He can invest part of his money at 8% interest
jeka94

Answer:

Amount invested at 8% is $9000 and amount invested at 18% is $21000.

Step-by-step explanation:

Let amount invested at 8% be x.

Let amount invested at 18% be y.

We get the 1st equation as:

x+y=30000       ........(1)

We get the second equation as:

0.08x+0.18(y)=0.15\times30000

=> 0.08x+0.18y=4500

or getting rid of the decimal by multiplying by 100 on both sides.

8x+18y=450000          ........(2)

Multiplying (1) by 8 and subtracting from (2) we get

10y=210000  

So, y = 21000

And x+21000=30000

x = 30000-21000

So, x = 9000

Therefore, amount invested at 8% is $9000 and amount invested at 18% is $21000.

6 0
3 years ago
1. Leo made a container to store his camping gear. The container is in the shape of a rectangular prism. The container has mater
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7 0
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justin wants to buy shorts that cost $35 and some shirts that are on sale for $22 each. He has $95 to spend. Write an inequality
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3 years ago
A nonprofit bank that provides loans to small entrepreneurs in developing countries is considering organizing a mandatory traini
pashok25 [27]

Answer:

Step-by-step explanation:

From the given information:

For trained subjects:

sample size n_1 = 1200

The sample mean x_1 = 789

For non-trained subjects:

Sample size n_2 = 1200

The sample mean = 632

For trained subjects, the proportion who repaid the loan is:

\hat p_1 = \dfrac{x_1}{n_1}

\hat p_1 = \dfrac{789}{1200}

\hat p_1 = 0.6575

For non-trained loan takers, the proportion who repaid the loan was:

\hat p_2 = \dfrac{x_2}{n_2}

\hat p_2 = \dfrac{632}{1200}

\hat p_2 = 0.5266

The confidence interval for the difference between the given proportion is:

= [ ( \hat p_1 - \hat p_2 ) - E \ , \ (\hat p_1 - \hat p_2 ) + E  ]

where;

Level of significance = 1 - C.I

= 1 - 0.95

= 0.05

Z - Critical value at  ∝ = 0.05 is 1.96

The Margin of Error (E) = Z_{\alpha/2} \times \sqrt{\dfrac{\hat p_1 (1- \hat p_1) }{n_1} + \dfrac{\hat p_2 (1- \hat p_2)}{n_2} }

=1.96 \times \sqrt{\dfrac{0.658 (1- 0.658) }{1200} + \dfrac{0.527 (1- 0.527)}{1200} }

= 1.96 \times \sqrt{\dfrac{0.658 (0.342) }{1200} + \dfrac{0.527 (0.473)}{1200} }

= 1.96 \times \sqrt{1.8753 \times 10^{-4}+2.07725833 \times 10^{-4} }

= 1.96 × 0.019881

≅ 0.039

The lower limit = ( \hat p_1 - \hat p_2) - E

= (0.658 - 0.527) - 0.0389

= 0.131  - 0.0389

= 0.092

The upper limit = ( \hat p_1 - \hat p_2) + E

= (0.658 - 0.527) + 0.0389

= 0.131  + 0.0389

= 0.167

Thus, 95% C.I for the difference between the proportion of trained and non-trianed loan takers who repaired the loan is:

=0.092 \le p_1-p_2 \le 0.167

For this study;

The null hypothesis is:

H_o : p_1 -p_2 = 0

The alternative hypothesis is:

H_a : p_1 -p_2 \ne 0

Since the C.I lie between (0.092, 0.17);

And the null hypothesis value does not lie within the interval (0.092, 0.17).

∴

we reject the null hypothesis H_o at ∝(0.05).

Conclusion: We conclude that there is enough evidence to claim that the proportion of trained and non-trained loan takers who repaired the loan are different.

8 0
3 years ago
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