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BabaBlast [244]
3 years ago
6

56.8 km equals how many miles

Mathematics
2 answers:
Alex3 years ago
8 0
56.8 km is 35.29 miles :)
lianna [129]3 years ago
5 0

Answer:

35.29 miles,hope this helps.

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Convert 72°F to °C (to the nearest tenth) using one of the following formulas: F = 9/5xC + 32 or C = (F - 32) x 5/9.
Fudgin [204]
First set up the equation:
C=(F-32)\times\frac{5}{9}
Then substitute in known values:
C=(72-32)\times\frac{5}{9}
and solve for C
C=40\times\frac{5}{9}
C=22.22
8 0
3 years ago
Read 2 more answers
Help its pi, ill give brainiest to the first answer show work please :)
saw5 [17]

Answer:

16m

8m

50.24m

200.96m^2

Step-by-step explanation:

13. 8*2=16

14. 8m

15. 2pi r = 2* pi* 8= 50.24

16. pi r^2 = pi * 8^2= 200.96m^2

8 0
3 years ago
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Does anybody know the answer?
denis23 [38]

Answer:

7/4

Step-by-step explanation:

4 0
3 years ago
How do I find the percent of budget?​
ch4aika [34]

Answer:

First, subtract the budgeted amount from the actual expense. If this expense was over budget, then the result will be positive.

Next, divide that number by the original budgeted amount and then multiply the result by 100 to get the percentage over budget. If your expenses were lower than your budgeted amount, then this number will be negative, describing the percentage under budget.

6 0
3 years ago
The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
Mariana [72]

Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

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

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0
3 years ago
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