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umka21 [38]
4 years ago
5

What is 5/16 expressed as a percent? Enter your answer in the box.

Mathematics
1 answer:
kenny6666 [7]4 years ago
4 0

Answer:

31.25 %

Step-by-step explanation:


The fraction bar also means divide:

5 / 16

=

5 Divided By 16

=

0.3125

0.3125 * 100

= 31.25

- I.A -

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Write an equation in slope-intercept form of the line with slope 3/4 that contains the point (-4,1)
marusya05 [52]

Answer:

y = 3/4x + 4

Step-by-step explanation:

to put it into slope intercept form, you need to find the y intercept or b becasue the slope intercept form is y = mx + b where b is the y intercetpt and m is the slope. To find b you insert the points you were given so 1 = -4(3/4) + b

which simplifies to 4 = b so then you can put in your slope and y intercept into slope intercept form

6 0
3 years ago
The car dealership decides to revise its commission program. Now the salesperson gets a flat commission of $60 for every $200 of
DedPeter [7]
60 is what percent of 200...
60/200 = x/100
200x = 6000
x = 6000/200
x = 30%...that means for every sale, the sales person gets a 30% commission

C(p) = 0.3p <==
4 0
4 years ago
Read 2 more answers
A rectangular prism has a length of 8 in., a width of 4 in., and a height of 2 1/4 in.
Alexus [3.1K]
It would need 288 cubes to fill up the prism.
7 0
3 years ago
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You are asked to do a study of shelters for abused and battered women to determine the necessary capacity in your city to provid
ohaa [14]

Answer:

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

274 shelters will be needed.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 250, \sigma = 75

If the city’s shelters have a capacity of 350, will that be enough places for abused women on 95% of all nights?

What is the percentile of 350?

This is the pvalue of Z when X = 350.

Z = \frac{X - \mu}{\sigma}

Z = \frac{250 - 150}{75}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082.

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

If not, what number of shelter openings will be needed?

The 95th percentile, which is X when Z = 1.645. So

Z = \frac{X - \mu}{\sigma}

1.645 = \frac{X - 150}{75}

X - 150 = 1.645*75

X = 274

274 shelters will be needed.

5 0
3 years ago
Exhibit B: A restaurant has tracked the number of meals served at lunch over the last four weeks. The data shows little in terms
Mekhanik [1.2K]

Answer:

Option E is correct.

The expected number of meals expected to be served on Wednesday in week 5 = 74.2

Step-by-step Explanation:

We will use the data from the four weeks to obtain the fraction of total days that number of meals served at lunch on a Wednesday take and then subsequently check the expected number of meals served at lunch the next Wednesday.

Week

Day 1 2 3 4 | Total

Sunday 40 35 39 43 | 157

Monday 54 55 51 59 | 219

Tuesday 61 60 65 64 | 250

Wednesday 72 77 78 69 | 296

Thursday 89 80 81 79 | 329

Friday 91 90 99 95 | 375

Saturday 80 82 81 83 | 326

Total number of meals served at lunch over the 4 weeks = (157+219+250+296+329+375+326) = 1952

Total number of meals served at lunch on Wednesdays over the 4 weeks = 296

Fraction of total number of meals served at lunch over four weeks that were served on Wednesdays = (296/1952) = 0.1516393443

Total number of meals expected to be served in week 5 = 490

Number of meals expected to be served on Wednesday in week 5 = 0.1516393443 × 490 = 74.3

Checking the options,

74.3 ≈ 74.2

Hence, the expected number of meals expected to be served on Wednesday in week 5 = 74.2

Hope this Helps!!!

8 0
3 years ago
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