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SSSSS [86.1K]
3 years ago
6

William cycles at a speed of 15 miles per hour. he cycles for 12 miles from home to school. if he increases his cycling speed by

5 miles per hour, how much faster will he reach his school.
Mathematics
1 answer:
Alja [10]3 years ago
3 0
12/15=4/5 original speed
12/20=3/5 increase by 5mph
He will reach 12 minutes earlier than he goes 15 miles per hour.
Correct me if I am wrong.
You might be interested in
a farmer uses 1/3 of his land to plant cassava 3/5 of the remaining land to plant Maize and the rest vegetables if vegetables co
Rzqust [24]

Answer:

37.5 acres.

Step-by-step explanation:

Fraction of his land  where vegetables are grown

= 1 - 1/3 - 2/3 *3/5

(after taking away 1/3 we get 2/3 and the farmer uses 3/5 of this 2/3 to grow Maize.

= 1 - 1/3 - 6/15              LCD of 3 and 15= 15  so we write:

= 1 - 5/15 - 6/15

= 1 - (5+6)/16

=  1 - 11/15

= 4/15.

4 /15 is equivalent to  10 acres

so  total  area (= 1)  =   10 / 4/15     Invert the 4/15 and multiply:

= 10 * 15/4

=  150/4

= 37.5 acres.

3 0
3 years ago
Lines a and b are parallel<br> What is the value of x<br> 5<br> 10<br> 35<br> 55
arsen [322]

Answer:

5

Step-by-step explanation:

3 0
3 years ago
Is Alexis drove her car for three hours and drove 150 miles and then drove her car for 350 miles in four hours what is her avera
goblinko [34]
Total miles:350+150=500
Total hours:4+3=7
Avg speed:500/7=71.429 mph
4 0
4 years ago
A car dealership is having a sale in which you pay 85% of the retail price for a new vehicle. keith is buying a vehicle with a r
sertanlavr [38]
The answer should be <span>4819.5  :)</span>
4 0
4 years ago
According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif
suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

P(0.79

3 0
3 years ago
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