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miskamm [114]
3 years ago
13

Please show your work

Mathematics
1 answer:
AURORKA [14]3 years ago
5 0
This is the expression 18k + 22 p - 10
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Julio Ernesto earns $15 an hour at his part-time job. Last week he worked 26 hours, and 45 minutes.. What was his gross pay for
zmey [24]

Answer:I need help on my question pls help me

Step-by-step explanation:

8 0
3 years ago
a salesperson receives a 2.5% commission on sales. what commission does the salesperson receive for $8000 in sales?
Gelneren [198K]
The sales person receives a $200 sales commission. To get the answer, you would convert the percentage to a decimal, in this case... 2.5%, would become 0.025. You would then multiply 0.025•8000 to get your answer of $200. Which means in total, the salesperson would receive $8200.
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3 years ago
A landscaper makes a $2,000 profit in a week when he services 25 lawns. He makes a $600 profit in a week when he services 11 law
SOVA2 [1]
2000 - 600 = 1400
-------------- 
25 - 11 = 14

100 dollars per 1 lawn. 

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He will earn $3,600.
8 0
3 years ago
Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. Th
Verizon [17]

Answer:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

Step-by-step explanation:

We know the following info:

\bar X_1 = 21.9 sample mean for group 1

\bar X_2 = 19.8 sample mean for group 2

s_1 = 3.4 sample standard deviation for group 1

s_2 = 3.5 sample standard deviation for group 2

n_1 = 200 sample size group 1

n_2 = 200 sample size group 2

We want to find a confidence interval for the difference of means and the correct formula to do this is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

Now we just need to find the critical value. The confidence level is 0.95 then the significance is 1-0.95 =0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df= n_1 +n_2 -2= 200+200-2= 398

The critical value for this case would be :t_{\alpha/2}=1.966  

And replacing into the confidence interval formula we got:

(21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422

(21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778

And we are 95% confident that the true difference means are between 1.422 \leq \mu_1 -\mu_2 \leq 2.778

5 0
3 years ago
Find the value of x for which / || m
Tresset [83]

Answer:

Step-by-step explanation:

7 0
3 years ago
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