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Reika [66]
3 years ago
9

Shania deposited $3,000 into a savings account tha

Mathematics
1 answer:
tia_tia [17]3 years ago
8 0
$4,950 because 6.5 of 3000 is 195 so multiply that by 10
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Please show me how to put 0.64 into a fraction
BigorU [14]

Answer:

64/100 is 0.64 into a fraction

8 0
3 years ago
HELP PLEASE!!!! I NEED HELP WITH GEOMETRY!!!
lesya692 [45]

Step-by-step explanation:

m < QRS = m < URS + m < QRU

30x - 4 = 60° + 15x - 4

15x = 60°

X = 4°

<u>m < QRS</u> = 30 * 4 - 4 = <u>116°</u>

8 0
3 years ago
For f(x) = 3x + 1 and g(x) = x2 – 6, find (f + g)(x)
serg [7]
(f+g)(x)=f(x)+g(x)=3x+1+x²-6=x²+3x-5

answer is D
4 0
3 years ago
The bases of a right prism are rhombi with area A = 44 cm2. The height of the prism is h=2.3 dm. Find the volume V of the prism.
Allushta [10]
The volume is 1012 cm³.

The volume of a right prism is found by multiplying the area of the base and the height.  First we need to convert the height to centimeters.

1 dm = 10 cm
2.3 dm = 2.3(10) = 23 cm

Now we have 44(23) = 1012 cm³
8 0
3 years ago
A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
nlexa [21]

Answer:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
3 years ago
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