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2 years ago

9

A first-aid booth is located 3/4 of the distance from point A to point B. At what marker on the path is the first-aid booth loca

ted

1 answer:

Anastaziya [24]2 years ago

3 0

3/4*8= 24/4= 6 this is 3/4 the distance between point a and b

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If the doctor’s office is 7 units from the zoo and 2 units from the park, which ordered pair could represent the location of the

s2008m [1.1K]

Answer:

(-4, -2) is the answer

Step-by-step explanation:

is you move 7 units down from the zoo you will see ur also 2 units from the park and the coordinates would be (-4, -2)

Hope this helps :)

6 0

2 years ago

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What is the LCM of 4,8,14?

Tamiku [17]

The LCM of 4,8,14 is 56

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M of 8 are 8,16,32,48,56,64
M of 14 are 14,28,42,56

I hope this helps you

5 0

3 years ago

richard spends 30% of his monthly budget on rent. His total monthly is 3,000. How much does Ricard spend on rent

Alexeev081 [22]

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3 years ago

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I’d appreciate any help please!!! 30 points

Aliun [14]

7. 3c+3=84
3c=81
c=27
So either 81/3 but if simplified its 27/1

8. The correct answer would be 55=31+t

8 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

3 years ago