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2 years ago

I need help I can't figure out how to do this. I have to find the value of v. 4/5(v−7)=2

Mathematics

1 answer:

Nina [5.8K]2 years ago

7 0

9.5 because 9.5-7 is 2.5. 2.5 times 4/5 is 10/5 which is equivalent to 2. 4/5(2.5-7)=2

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How much pure acid is in 780 milliliters of a 12% solution?

GarryVolchara [31]

Answer:

93.6 ml

Step-by-step explanation:

12% of that 780 ml is pure acid: 0.12(780 ml) = 93.6 ml

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2 years ago

6<br> The quotient of 20 and 47.

Semenov [28]

Answer:

Step-by-step explanation:

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2 years ago

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Find the surface area

Brut [27]

Answer:

Step-by-step explanation:

area of top=7×6=42 yd²

area of bottom=3×6=18 yd²

area of two rectangles=2[4×6]=48 yd²

area of two trapezoids=2[1/2(7+3)×3.5]=35 yd²

total surface area=42+18+48+35=143 yd²

7 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

from 1970 to 1980 the consumer price index increased from 38.8 to 82.4. If a container of onion dip cost $0.48 in 1970 and the p

Vikki [24]

We need to find the percent increase from 38.8 to 82.4, so

x * 38.8 = 82.4
x = 82.4/38.8
x = 2.1237

Multiply by 100 to convert to percentage (move decimal 2 places to the right)

2.1237 = 212.37%

Now we can apply this percentage increase to the onion dip...

0.48 x 212.37% =
0.48 x 2.1237 = 1.019

Round to the nearest cent $1.02 is the amount of increase. Now add it to the original price to get today's price...

$1.02 + .48 = $1.50

5 0

2 years ago

Read 2 more answers