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2 years ago

I need help asap!!!

Mathematics

1 answer:

gizmo_the_mogwai [7]2 years ago

5 0

Answer:

x = 9

Step-by-step explanation:

-5(1 + 1x) = -50

-5 + -5x = -50

-5 + -5x + 5 = -5x

-50 + 5 = -45

-5x = -45

-5x / -5 = x

-45 / -5 = 9

x = 9

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DUE IN 5 MINS PLEASE HELP

Maksim231197 [3]

Answer:

Step-by-step explanation:

you can pack one pack in 20 min

there are three 20 min in one hour

multiply three times the four hours and you get 12

5 0

3 years ago

In 2000, the number of students enrolled at Arlington Country Day School was 823 students. In 2010, the population was approxima

vivado [14]

705 - 823 = -118

-118/823 = -0.143 = 14.3% decrease

5 0

3 years ago

Read 2 more answers

The national average sat score (for verbal and math) is 1028. if we assume a normal distribution with standard deviation 92, wha

elena55 [62]

Let X be the national sat score. X follows normal distribution with mean μ =1028, standard deviation σ = 92

The 90th percentile score is nothing but the x value for which area below x is 90%.

To find 90th percentile we will find find z score such that probability below z is 0.9

P(Z <z) = 0.9

Using excel function to find z score corresponding to probability 0.9 is

z = NORM.S.INV(0.9) = 1.28

z =1.28

Now convert z score into x value using the formula

x = z *σ + μ

x = 1.28 * 92 + 1028

x = 1145.76

The 90th percentile score value is 1145.76

The probability that randomly selected score exceeds 1200 is

P(X > 1200)

Z score corresponding to x=1200 is

z = $\frac{x - mean}{standard deviation}$

z = $\frac{1200-1028}{92}$

z = 1.8695 ~ 1.87

P(Z > 1.87 ) = 1 - P(Z < 1.87)

Using z-score table to find probability z < 1.87

P(Z < 1.87) = 0.9693

P(Z > 1.87) = 1 - 0.9693

P(Z > 1.87) = 0.0307

The probability that a randomly selected score exceeds 1200 is 0.0307

5 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

In a class of 40 students, everyone has either a pierced nose or a pierced ear. The professor asks everyone with a pierced nose

viktelen [127]

Answer:

3 students

Step-by-step explanation:

If everyone in the class has either a pierced nose or ear, we just simply have to add up the total number of hands raised and minus the number of students in the class.

35+8=43

43-40=3

3 students have both a pierced nose and pierced ear.

6 0

3 years ago