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2 years ago

HELP ASAP ILL GIVE BRAINLIEST!!! Dividing mixed numbers^^

Mathematics

1 answer:

saul85 [17]2 years ago

6 0

1) 199/30
2) 95/12
3) 47/10
4) 103/15
5) 83/10
6) 221/30

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An automobile insurance company divides customers into three categories: good risks, medium risks, and poor risks. Assume that o

Leokris [45]

Answer:

0.60

Step-by-step explanation:

Probability that the customer is not a poor risk = 1 - probability that the customer is a poor risk

Firstly, let’s calculate the probability of being a poor risk.

From the given data the number of poor risks = 14229-7362-1190 = 5677

So the probability of being a poor risk = 5677/14229 = 0.399

Thus, the probability that the customer is not a poor risk = 1-0.399 = 0.601 which to 2 decimal places = 0.60

6 0

3 years ago

Ella is a landscape photographer. One weekend at her gallery she sells a total of 52 prints for a total of $2,975. A small paint

rosijanka [135]

<h2>Answer:37 paintings of $50 and 15 paintings of $75</h2>

Step-by-step explanation:

Let $x$ be the number of paintings Ella sells for $ $50$ .

Let $y$ be the number of paintings Ella sells for $ $75$ .

Profit made through $ $50$ paintings is $50x$

Profit made through $ $75$ paintings is $75y$

So,total profit is given by $50x+75y$

It is given that total profit is $ $2975$

So, $50x+75y=2975$ ..(i)

Given that the total number of prints is $52$

So, $x+y=52$ ..(ii)

using (i) and (ii),

$50x+75(52-x)=2975\\50x+3900-75x=2975\\25x=925\\x=37$

$y=52-x=52-37=15$

7 0

3 years ago

Read 2 more answers

In order to estimate the average time spent on the computer terminals per student at a local university, data were collected fro

VladimirAG [237]

Answer:

Margin of error for a 95% of confidence intervals is 0.261

Step-by-step explanation:

Step1:-

Sample n = 81 business students over a one-week period.

Given the population standard deviation is 1.2 hours

Confidence level of significance = 0.95

Zₐ = 1.96

Margin of error (M.E) = $\frac{Z_{\alpha }S.D }{\sqrt{n} }$

Given n=81 , σ =1.2 and Zₐ = 1.96

Step2:-

 $Margin of error (M.E) = \frac{Z_{\alpha }S.D }{\sqrt{n} }$ 

 $Margin of error (M.E) = \frac{1.96(1.2) }{\sqrt{81} }$ 

On calculating , we get

Margin of error = 0.261

Conclusion:-

Margin of error for a 95% of confidence intervals is 0.261

4 0

3 years ago

You have to solve the problems starting at the "begin" and make your way up to the "end". The correct answer is one of the probl

Thepotemich [5.8K]

Begin
3x-4=14
3x=18
x=6

.5n=1.25
n=2.5

x+1=2x-2
x+3=2x
3=x

x+3=10
x=7

6x-13=-11
6x=2
x=1/3

p=4-p
2p=4
p=2

End

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2 years ago

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. For the function given, state the starting point for a sample period:

Aneli [31]

I think the answer is π?

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3 years ago