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Feliz [49]
3 years ago
15

If you drive at a speed of 50 mph how many hours will it take to drive 1340 miles

Mathematics
1 answer:
timama [110]3 years ago
4 0

It would take you 26.8 hours to drive 1340 miles... I think.

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Complete the table thanks in advance
inna [77]
Y = 1/2 x

for x = 0 → y = 1/2 · 0 = 0
for x = 2 → y = 1/2 · 2 = 1


   x      | 0 | 2 |
y=1/2x| 0 | 1 |


y = x + 3

for x = 0 → y = 0 + 3 = 3
for x = -3 → y = -3 + 3 = 0

    x    | 0 | 3 |
y=x+3|-3 | 0 |
6 0
3 years ago
1 Kg = 2.2046 pounds what will be the answer in round to nearest tenth 4550 pounds = ______ kg
Harrizon [31]

Answer:

2063.9 kg

Step-by-step explanation:

Given,

<u>1 kg = 2.2046 pounds</u>

This can also be written as:

<u>1 pound = 1/2.2046 kg</u>

We have to calculate the value of 4550 pounds in kg up to nearest 10th

Thus,

4550 pounds= \frac{4550}{2.2046} kg

Solving the above equation, we get:

4550 pounds = 2063.8664 kg

Rounding the above result to nearest tenth as:

<u>4550 pounds = 2063.9 kg</u>

6 0
3 years ago
Here is a list of numbers:<br> 3,16, 7, 19, 1, 12, 3,9,16,12<br> State the median.
fomenos

Answer:

10.5

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is 10a - 5b if a = 25 and b = 3​
AURORKA [14]

Answer:

\boxed{235}

Step-by-step explanation:

<em>Hey there!</em>

Well given that,

a = 25

b = 3

10(25) - 5(3)

250 - 15

= 235

<em>Hope this helps :)</em>

6 0
3 years ago
Read 2 more answers
On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to
NemiM [27]

Answer:

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.2622

The margin of error is:

M = T*s = 2.2622*0.3 = 0.68

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 98.44 - 0.68 = 97.76 ºF

The upper end of the interval is the sample mean added to M. So it is 98.44 + 0.68 = 99.12 ºF

The 95% confidence interval for the mean of all body temperatures is between 97.76 ºF and 99.12 ºF

4 0
3 years ago
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