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3 years ago

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It takes Ms. Brown 512 hours to hike 11 miles. At this rate, how long will it take her to hike the remaining 7 miles to her camp

site?

1 answer:

Rudiy273 years ago

6 0

Answer:

325 hours

Step-by-step explanation:

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Which is the correct way to represent 0.0035 kg by using scientific notation? –3.5 103 kg –3.5 10–3 kg 3.5 10–3 kg 3.5 103 kg?

Vlada [557]

It would be

3.5 × 10⁻³ kg.

The decimal point must be moved 3 places to the right in order to have it behind the first non-zero digit; this gives us the exponent of 3, and since we are moving the decimal to the right, it is a negative exponent.

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Nezzie is shopping for items for her daycare because the store is having a sale where everything in the store is 10% off. Nezzie

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Soo 100.26-15% off = $85.25 and 59.99 you need to crank it up 1 so $60-10% off=$54 last add $85.25+$54=$129.25

So no she is not right

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4 years ago

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Please answer the following question. Also please answer if you actually know it.

Iteru [2.4K]

Given : f(x)= 3|x-2| -5

f(x) is translated 3 units down and 4 units to the left

If any function is translated down then we subtract the units at the end

If any function is translated left then we add the units with x inside the absolute sign

f(x)= 3|x-2| -5

f(x) is translated 3 units down

subtract 3 at the end, so f(x) becomes

f(x)= 3|x-2| -5 -3

f(x) is translated 4 units to the left

Add 4 with x inside the absolute sign, f(x) becomes

f(x)= 3|x-2 + 4| -5 -3

We simplify it and replace f(x) by g(x)

g(x) = 3|x + 2| - 8

a= 3, h = -2 , k = -8

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

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3 years ago

Solve for y ..............

lions [1.4K]

Answer:

Im not gonna explain i will give you the answer that i came up with I currently have 60 right now hope this is right

Step-by-step explanation:

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3 years ago