All categories

2 years ago

2n-7/3 = 15 solve for n

1 answer:

5 0

$2n- \frac{7}{3} =15$
$2n- \frac{7}{3} =15$ common denominator is 3
$3(2n- \frac{7}{3} )=3*15$ distributhing 3 through parantheses
$3*2n- \frac{3*7}{3}=3*15$
$3*2n-7=3*15$
$6n-7=45$ add7
$6n=45+7$
$6n=52$ divide by 6
$n= \frac{52}{6}$ simplify
$n=8 \frac{2}{3}$ ⇒ $\frac{26}{3}$

You might be interested in

Please help please help please helppppppppppppppppppp

Amiraneli [1.4K]

Answer:

600

Step-by-step explanation:

3 0

2 years ago

Seven subtracted from three times a number is -4

Fynjy0 [20]

X - the number

3x - 7 = -4 |add 7 to both sides

3x = 3 |divide both sides by 3

x = 1

7 0

3 years ago

COSS SENSE-MAKING The sides of square ABCD are extended

dsp73

Answer: 9 in

Step-by-step explanation:

5 0

2 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

Through (-4,6) and parallel to y=-3x+4

liraira [26]

Y=mx+b
parallel lines will have the same slope.

using the points (-4,6) and the slope for the parallel line -3 we can find b (the y intercept) for the line.

6 = (-3)(-4) + b
6 = 12 + b
-12 + 6 = -12 + 12 +b
-6 = b
so the line that runs through points (-4,6) and parallel to y= -3x+4 is:
y = -3x -6

7 0

2 years ago