What are the factors of 35

Half an hour after leaving her house Beverly was 20 miles from home. At this point she merged onto the highway and was traveling

N76 [4]

Answer:

1. 45.71 miles

2. 10 miles

3. 89.43 miles

4. 0.79 hr

5. 0.025 hr

Step-by-step explanation:

speed = distance/time

After 4 hrs Beverly was 200 miles from home , however she had spent 0.5hr travelling for 20 miles prior to getting to the highway.

Therefore;

Constant speed before Highway = 20/0.5 = 40 miles/hr

Constant speed at Highway = $\frac{(200-20)}{(4-0.5)}$ = 51.43 miles/hr

1. in 1 hr beverly will have travelled;

⇒ (20 miles in 0.5 hr ) + (0.5 hr at speed 51.43 miles/hr)

⇒ 20 + (0.5 x 51.43) = 45.71 miles

2. In 0.25 hr beverly had not reached the highway

⇒ distance = speed x time

⇒ distance = 40 x 0.25 = 10 miles

3. in 1.85 hrs beverly will have travvelld;

⇒ (20 miles in 0.5 hr ) + (1.35 hr at speed 51.43 miles/hr)

⇒ 20 + 69.43 = 89.43 miles

4. Time taken to get to 35 miles

⇒ (20 miles in 0.5 hr ) + (15 miles at speed 51.43 miles/hr)

⇒ 0.5hr + (15/51.43) hr = 0.79 hr

5. Time taken to travel 1 mile

⇒ time = distance / speed = 1 mile / 40 miles/hr

⇒ time = 0.025 hr

8 0

3 years ago

Henry ran five races, each of which was a different positive integer number of

dolphi86 [110]

Answer:

10 miles.

Step-by-step explanation:

Let x be the number of miles on Henry's longest race.

We have been given that Henry ran five races, each of which was a different positive integer number of miles.

We can set an equation for the average of races as:

$\frac{\text{The sum distances of 5 races}}{5}=4$

As distance covered in each race is a different positive integer, so let his first four races be 1, 2, 3, 4.

Now let us substitute the distances of 5 races as:

$\frac{1+2+3+4+x}{5}=4$

$\frac{10+x}{5}=4$

Let us multiply both sides of our equation by 5.

$\frac{10+x}{5}*5=4*5$

$10+x=20$

Let us subtract 10 from both sides of our equation.

$10-10+x=20-10$

$x=10$

Therefore, the maximum possible distance of Henry's longest race is 10 miles.

7 0

3 years ago

A professor pays 25 cents for each blackboard error made in lecture to the student who pointsout the error. In a career ofnyears

marta [7]

Answer:

(a) The probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b) n = 28.09

Step-by-step explanation:

The random variable Yₙ is defined as the total numbers of dollars paid in n years.

It is provided that Yₙ can be approximated by a Gaussian distribution, also known as Normal distribution.

The mean and standard deviation of Yₙ are:

$\mu_{Y_{n}}=40n\\\sigma_{Y_{n}}=\sqrt{100n}$

(a)

For n = 20 the mean and standard deviation of Y₂₀ are:

$\mu_{Y_{n}}=40n=40\times20=800\\\sigma_{Y_{n}}=\sqrt{100n}=\sqrt{100\times20}=44.72\\$

Compute the probability that Y₂₀ exceeds 1000 as follows:

$P(Y_{n}>1000)=P(\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}>\frac{1000-800}{44.72})\\=P(Z> 4.47)\\=1-P(Z$

**Use a z table for probability.

Thus, the probability that Y₂₀ exceeds 1000 is 3.91 × 10⁻⁶.

(b)

It is provided that P (Yₙ > 1000) > 0.99.

$P(Y_{n}>1000)=0.99\\1-P(Y_{n}$

The value of z for which P (Z < z) = 0.01 is 2.33.

Compute the value of n as follows:

$z=\frac{Y_{n}-\mu_{Y_{n}}}{\sigma_{Y_{n}}}\\2.33=\frac{1000-40n}{\sqrt{100n}}\\2.33=\frac{100}{\sqrt{n}}-4\sqrt{n} \\2.33=\frac{100-4n}{\sqrt{n}} \\5.4289=\frac{(100-4n)^{2}}{n}\\5.4289=\frac{10000+16n^{2}-800n}{n}\\5.4289n=10000+16n^{2}-800n\\16n^{2}-805.4289n+10000=0$