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3 years ago

write word problems that represents each way you can use a remainder in a division problem. Include solutions

1 answer:

4 0

Jack has 60 toys. He wants to store all of them in boxes. Each box can hold 9 toys. How many boxes does Jack need?

The key word is 'all'. There is a remainder because 9 doesn't go into 60 evenly. He wants to store all the toys, so he needs an extra box for the leftover toys.

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What is the simplified form of 12z^2-7z-12/3z^2+2z-8? ...?

gavmur [86]

It should go like this:

(4z + 3) (3z - 4) / (3z - 4) (z + 2)

Then you just cancel out (3z -4) and (3z -4), and the final simplified form of this polynomial is:

(4z +3) / (z + 2)

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3 years ago

What is 5 divided by 1/4

velikii [3]

The answer is 20.

Hope this helps!! May I have brainliest?

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3 years ago

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Is x^2-3x-23 polynomial

umka2103 [35]

Answer:

Yes

Step-by-step explanation:

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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

Can someone think of a poem which starts with "Today I feel ...."<br>pls

GuDViN [60]

Yes, I shall give you an idea of how to start it. :)

Today I feel glad...
For all the friends I have had...
They gave me a pleasure...
For all that I treasure...
Special and dear to my heart.

Ta da! :D Hope you like it

7 0

3 years ago

Read 2 more answers