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3 years ago

I need help whats the answer ?

Mathematics

2 answers:

ladessa [460]3 years ago

6 0

Here, we're looking for the unit rate, or how many words James can type in 1 minute.

Let's use 300 words in 10 minutes to find words per minute.

300 words / 10 minutes = 30 words per minute

Hope this helps!

MrMuchimi3 years ago

6 0

Answer:

30 words per minute

Step-by-step explanation:

Take the number of words and divide by the number of minutes

150/5 = 30

300/10 =30

450/15 = 30

600/20 = 30

30 words per minute

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Blake is paid $8.75 an hour. He worked 6 hours and spent 1 5 of his pay on a book. How much money did he have after he paid for

jek_recluse [69]

Answer: $42

Step-by-step explanation:

His total pay is $52.50 = 8.47x6

if he spent 1/5 then he has 4/5 left. 4/5 times 52.50 = 4(10.50 = $42

3 0

3 years ago

Rebecca answered 42 of 60 questions correctly on a test which method should he she used to find the percent of question she answ

MakcuM [25]

She answered 70% questions correctly

Further explanation:

Percetages are used to calculae the success or failure in tests or exams usually.

Given

$Total\ questions= 60\\The\ questions\ attempted\ correctly=42$

The percentage will be calculated as follows:

$Percentage\ of\ correctly\ answered\ questions=\frac{Correct\ questions}{Total\ questions}*100\\= \frac{42}{60}*100\\=70\%$

She answered 70% questions correctly

Keywords: Percentage, Success rate

Learn more about percentage at:

#LearnwithBrainly

3 0

3 years ago

Read 2 more answers

37.5% of what number is 96

marshall27 [118]

Okay the way to find this out is to multiply 96 by .375 (.375 is the decimal version of 37.5%) so 96×.375=36. So 37.5% of 96 is 36 ANSWER: 36

8 0

3 years ago

Identify the slope.Nadia's pool already has 1000 gallons of water when she begins to fill it. It fills at a rate of 22 gallons p

Nataly_w [17]

Y=22x+100
22 would be the slope

3 0

3 years ago

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming

Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by $\theta$ and in order to construct a confidence interval we can use the following formula:

$\hat \theta \pm ME(\hat \theta)$

Where $\hat \theta$ is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

$ME = z_{\alpha} SE$

Where $z_{\alpha/2}$ is a quantile from the normal standard distribution that accumulates $\alpha/2$ of the area on each tail of the distribution. And $SE$ represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

$SE= \frac{\hat p (1-\hat p)}{n}$

And if our parameter of interest is the sample mean the standard error is given by:

$SE = \frac{s}{\sqrt{n}}$

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

$SE = f(n)$

And since the margin of error is a multiple of the standard error we have that $ME = f(n)$

Now if we find the width for a confidence interval we got this:

$Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]$

$Width = 2 ME (\hat \theta)$

And we can express this as:

$Width =2 f(n)$

And we can define the function $f(n) = \frac{1}{\sqrt{n}}$ since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

$Width_i= 2 \frac{1}{\sqrt{n}}$

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

$Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i$

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0

4 years ago