If a person ran 1/2 of a mile in 1/10 of an hour. How far will the person run in one (1) hour?

PLEASE HELP ME WITH THIS PROBLEM ASAP

lana [24]

Answer:

15x+5+22x+4+60=180

37x=180-60-5

37x=115

x=115/37

x=3

angle b

22*3+4=70°

angle c

15*3+5=50°

have a nice day . hope this helps

7 0

3 years ago

joes family has compleated 60% of a road trip. if they have traveld 30 miles, how far is the full road trip?

Tems11 [23]

The full road trip is 50 miles.

Step-by-step explanation:

Given,

Distance traveled by Joe's family = 30 miles

This represents 60% of the total distance to be traveled.

Let,

x be the total distance.

60% of x = 30

$\frac{60}{100}*x=30\\0.6x=30$

Dividing both sides by 0.6

$\frac{0.6x}{0.6}=\frac{30}{0.6}\\x=50\ miles$

The full road trip is 50 miles.

Keywords: percentage, division

Learn more about percentages at:

#LearnwithBrainly

4 0

3 years ago

What is the greatest common factor of 6a, 8a^2, and 14?

Sergio [31]

Answer:

8a is the common favor of 6a

Step-by-step explanation:

hope this helps!!

8 0

3 years ago

Read 2 more answers

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

professor190 [17]

Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0

3 years ago

There are 75 students performing in the holiday concert. Of the students, 30 are boys. What percent of the students are boys?

viva [34]

Answer:

Total students = 75

Boys = 30

<h3>Percentage</h3>

$\sf \frac{30}{75} \times 100\% \\ \sf= \frac{10}{25} \times 100\% \\ \sf = \boxed{ \bold{ \red{40\%}}}$

Hope it helps!

8 0

3 years ago