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11111nata11111 [884]
3 years ago
9

Select the correct answer. Chloe enjoys her math classes, and she would like to find a career that will allow her to continue to

use her math skills. Which career would be a good fit for her?
a. accountant
b. purchasing agent
c. commercial carpenter
d. human resources manager
Mathematics
2 answers:
Arte-miy333 [17]3 years ago
6 0

It would have to be an Accountant.

gladu [14]3 years ago
3 0

Answer:

<h2>a. accountant</h2>

Step-by-step explanation:

According to the proble, Chloe ENJOYS her math classes. Also, she has math skills. These two clues indicate that Chloe shouls study a carreer related with math, because she enjoys it.

Therefore, the right answer is a. accountant, because an accountant need to have math skills as tools to the daily activity.

An accountant provides financial information that helps managers and directors to make important choices for the company. That financial information is an analysis made by using math skills.

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vagabundo [1.1K]

Answer:

Both are circle equations. One is for area, one is for circumference.

Step-by-step explanation:

Both are circle equations. One is for area, one is for circumference.

4 0
3 years ago
A pack of red pens can be divided into equal shares among 3, 5, 6 and 9 teachers with no pen left over?
Ugo [173]

Answer:

90 red pens

Step-by-step explanation:

we need to find the least common multiple of the numbers 3 , 5 , 6 , 9

6 = 2 × 3

9 = 3²

then

LCM(3,5,6,9) = 2×5×3² = 2×5×9 = 90

The number 90 can be divided into 3 ,5 ,6 ,9 with no remainder

Therefore the pack must have 90 pens or a multiple of 90

For example 180 , 270 ...

7 0
2 years ago
Find the slope of the line through (-9,-10) and (-2,-5)
valentina_108 [34]
Slope = Change in y direction / Change in x direction

m =  \frac{y_2-y_1}{x_2-x_1} =  \frac{-5-(-10)}{-2-(-9)} =  \frac{5}{7}


5 0
3 years ago
What integer $x$ satisfies $\frac{1}{4}&lt;\frac{x}{7}&lt;\frac{1}{3}$?
Zarrin [17]

If 1/4 < x/7 < 1/3, then

7/4 < x < 7/3

1.75 < x < 2.333…

so that x = 2.

6 0
2 years ago
The mean time required to repair breakdowns of a certain copying machine is 93 minutes. The company which manufactures the machi
Sonja [21]

Answer:

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

p_v =P(t_{(72)}  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=88.8 represent the sample mean

s=26.6 represent the sample standard deviation for the sample  

n=73 sample size  

\mu_o =93 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean i lower than 93 min, the system of hypothesis would be:  

Null hypothesis:\mu \geq 93  

Alternative hypothesis:\mu < 93  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{88.8-93}{\frac{26.6}{\sqrt{73}}}=-1.349    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=73-1=72  

Since is a one side test the p value would be:  

p_v =P(t_{(72)}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, and we can't concluce that the true mean is less than 93 min at 5% of signficance.  

5 0
3 years ago
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