23 times 58 divied by 23 minus 8 plus 5 times 1 dived by 7

10PTS pls help me on this math problem and pls explain how you did it.

erastovalidia [21]

Answer:

y = 8

Step-by-step explanation:

Line1 with points (4, 1) and (0, 0):

slope = (y₂ - y₁) / (x₂ - x₁)

= (0 - 1) / (0 - 4)

= 1/4

Perpendicular line is opposite inverse so original slope 1/4 becomes -4/1 or simply -4.

Line2 with points (1, 0) and (-1, y):

slope = (y₂ - y₁) / (x₂ - x₁)

= (y - 0) / (-1 - 1)

= y / -2

y must equal to 8 for the slope to be -4 as stated above.

4 0

3 years ago

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bank a charges a monthly service fee of $2.00 and a per check fee of 0.05, while bank B charges a monthly service fee of $1.50 a

saw5 [17]

2 + .05x = 1.5 +.1x .5 = .05x x =10 is numbere when costs are equal So the 11th check makes Bank A cheaper

3 0

3 years ago

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What is a function? *

Lera25 [3.4K]

Answer:

A function is a process or a relation that associates each element x of a set X, the domain of the function, to a single element y of another set Y (possibly the same set), the codomain of the function.

Step-by-step explanation:

hope this helps

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3 years ago

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The edge of each cube used to build this rectangular prism is 13

kolbaska11 [484]

Answer:

$\frac{3}{3} * \frac{2}{3} *\frac{3}{3} =\frac{2}{3} \ in^3$

====================================================

Step-by-step explanation:

The question is missing the attached figure.

The edge of each cube used to build this rectangular prism is 1/3 inch long. Based on the attached figure.

the volume of the prism = length × height × width

length = $3 * \frac{1}{3} = \frac{3}{3}$

height = $3 * \frac{1}{3} = \frac{3}{3}$

width = $2 * \frac{1}{3}= \frac{2}{3}$

So the volume of the prism = $\frac{3}{3} *\frac{3}{3} *\frac{2}{3} =\frac{2}{3} \ in^{3}$

∴ The equation that shows the volume of the prism, in cubic inches

= $\frac{3}{3} * \frac{2}{3} *\frac{3}{3} =\frac{2}{3} \ in^3$

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3 years ago

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Suppose you are working in an insurance company as a statistician. Your manager asked you to check police records of car acciden

pochemuha

Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ~ N(0,1)

where, $\hat p$ = sample proportion teenage drivers = $\frac{120}{576}$ = 0.21

n = sample of accidents = 576

p = population percentage of all car accidents

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ]

= [ $0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}$ , $0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}$ ]

= [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

Let p = percentage of teenagers who where involved in car accidents

So, Null Hypothesis, $H_0$ : p = 18% {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, $H_A$ : p $\neq$ 18% {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ~ N(0,1)

where, $\hat p$ = sample proportion teenage drivers = $\frac{120}{576}$ = 0.21

n = sample of accidents = 576

So, test statistics = $\frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}$

= 1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

4 0

3 years ago