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2 years ago

I need to find both of the solutions to the equation 100+(n-2)^ = 149

Mathematics

1 answer:

rodikova [14]2 years ago

5 0

Answer:

Step-by-step explanation:

hello :

100+(n-2)² = 149

100-100+(n-2)² = 149-100

(n-2)² = 49

(n-2)² - 49 =0 but 49=7²

(n-2)² - 7² =0 use identity : a²-b²=(a-b)(a+b)

(n-2-7)(n-2+7)=0

(n-9)(n+5)=0

n-9=0 or n+5=0

n=9 or n=-5

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A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

2 years ago

A salesman’s commission on a $150,000 sale is $3,000. What percent is his commission?

Elenna [48]

Answer:

The commission percentage is 2%

Step-by-step explanation:

To find the percentage of a commission, start by dividing the commission amount by the total amount sold.

3,000/150,000 = .02 = 2%

5 0

2 years ago

What is (0.5x + 15) + (8.2x- 16.6)

nadya68 [22]

1.6 I think I’m sorry if it’s incorrect

3 0

2 years ago

Daniel and his children went into a grocery store and he bought $10.15 worth of apples and bananas. Each apple costs $1.75 and e

guapka [62]

Answer: Daniel bought 3 apples and 7 bananas.

Step-by-step explanation:

Let x represent the number of apples that Daniel bought.

Let y represent the number of bananas that Daniel bought.

He bought a total of 10 apples and bananas altogether. This means that

x + y = 10

Daniel and his children went into a grocery store and he bought $10.15 worth of apples and bananas. Each apple costs $1.75 and each banana costs $0.70. This means that

1.75x + 0.7y = 10.15 - - - - - - - - - - - 1

Substituting x = 10 - y into equation 1, it becomes

1.75(10 - y) + 0.7y = 10.15

17.5 - 1.75y + 0.7y = 10.15

- 1.75y + 0.7y = 10.15 - 17.5

- 1.05y = - 7.35

y = - 7.35/- 1.05

y = 7

x = 10 - y = 10 - 7

x = 3

4 0

2 years ago

Tim was given a large bag of sweets and ate one third of the sweets before stopping as he was feeling sick. The next day he ate

mars1129 [50]

Answer: In the beginning he was given 27 sweets.

Step-by-step explanation: The most logical thing to do is to solve it backwards, that is, from what he had at the end of the third day up till the beginning of the first day.

On the third day he ate one-third and had 8 sweets left over. To determine how many he started with on the third day, let the total on day three be called a. If one-third of a is eaten, then the left over which is two-thirds is 8. That is;

8/a = 2/3

By cross multiplication we now have

8 x 3 = 2a

24/2 = a

a = 12

Let the number of sweets he had on day two be called b. If he ate one-third of b and he had 12 left over, then the two-thirds left over is 12 and we now have;

12/b = 2/3

By cross multiplication we now have

12 x 3 = 2b

36 = 2b

36/2 = b

b = 18

Let the number of sweets he had on day one be called x. If he ate one-third of x and he had 18 left over, then the two-thirds left over is 18, and we now have;

18/x = 2/3

By cross multiplication we now have

18 x 3 = 2x

54 = 2x

x = 27

Therefore Tim was given 27 sweets at the beginning.

3 0

2 years ago