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Alexus [3.1K]
3 years ago
9

An animal shelter recently held an adoption event and found homes for 48 of their shelter animals. If they found home for 80% of

the animals during the event, how many animals were in the shelter before the event?
Mathematics
1 answer:
Elden [556K]3 years ago
5 0
There were 60 animals before the event.
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Help pls and ty :D !!​
xxMikexx [17]

Try this suggested option, answers are marked with colour.

4 0
2 years ago
At the start of the month, the value of an investment was $73.42. By the end of the month, the value of the investment changed b
Aleonysh [2.5K]

Answer:

0.8%.

Step-by-step explanation:

Given:

At the start of the month, the value of an investment was $73.42.

By the end of the month, the value of the investment changed by a loss of $13.53.

Question asked:

What was the value, in dollars, of the investment at the end of the month?

What was the percent loss?

Solution:

At the start of the month, the value of an investment = $73.42.

Loss = $13.53.

<u><em>The value, of the investment at the end of the month = The value of an investment at the start of the month - loss</em></u>

The value, of the investment at the end of the month = $73.42 - $13.53

                                                                                         = $59.89

Thus, the value, of the investment at the end of the month is $59.89.

Now, we will find percent loss:-

Percent\ loss = \frac{Loss}{value\ of \ investment\ at \ start}

                    =\frac{59.89}{73.42} \\=0.81\%

Thus, percent loss on investment by the end of the month is 0.8%.

7 0
3 years ago
A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students a
Stella [2.4K]

Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.7291

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample. For this question, we have s = 5, n = 20. So

M = T\frac{s}{\sqrt{n}}

M = 1.7291\frac{5}{\sqrt{20}}

M = 1.933

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

8 0
3 years ago
PLZ HELP 30points and brainlist
MariettaO [177]
The answer is b so good luck and i hope that helps
4 0
3 years ago
Read 2 more answers
A random sample of 250 juniors majoring in psychology or communication at a large university is selected. These students are ask
fenix001 [56]

Solution :

It is given that :

Number of students in a random sample majoring in communication or psychology at an university = 250

Total number of students majoring in psychology = 100

Number of students majoring in psychology  those who are happy = 80

So number of students majoring in psychology  those who are not happy = 20

Total number of students majoring in communication = 250 - 100 = 150

Number of students majoring in communication  those who are happy = 115

So number of students majoring in psychology  those who are not happy = 35

a). Probability of the students happy with their major choices are

    $=\frac{80+115}{250}$

    = 0.78

b). Psychology major $=\frac{80+20}{250}$

                                   = 0.4

c). Probability of the students who are happy with the communication as the choice of major = $\frac{115}{250}$ = 0.46

d). Students unhappy with their choice of major given that the student is psychology major = $\frac{20}{250}$  =  0.018

8 0
3 years ago
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