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3 years ago

Jamie ran 18.6 miles in 3 hours. How far can she run in a given amount of time, if she runs at a constant rate?

2 answers:

7 0

For the graph just put a dot where 6.2 miles and one hour intersect and then where 12.4 miles and two hours and then where 18.6 miles in three hours intersect and then use a ruler to draw the line connecting them.

Irina-Kira [14]3 years ago

4 0

Answer:

6.2 per hour

Step-by-step explanation:

Could only help you with b sorry pal

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Solve the equation<br><br> 8^7x=4,096

natali 33 [55]

The answer to the question above is 2.4

6 0

4 years ago

Read 2 more answers

The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years. a. Find

Anvisha [2.4K]

Answer

a. 0.856

b. 0.78071

c. It is not unusual

d. 13.65 years old

Step-by-step explanation:

The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years.

We solve this question using z score formula:

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a. Find the probability that a cat will live to be older than 14 years.

For x > 14 years

z = 14 - 15.7/1.6

z = -1.0625

Probability value from Z-Table:

P(x<14) = 0.144

P(x>14) = 1 - P(x<14) = 0.856

b. Find the probability that a cat will live between 14 and 18 years.

For x = 14 years

z = 14 - 15.7/1.6

z = -1.0625

Probability value from Z-Table:

P(x = 14) = 0.144

For x = 18 years

z = 18 - 15.7/1.6

z= 1.4375

Probability value from Z-Table:

P(x = 18) = 0.92471

The probability that a cat will live between 14 and 18 years is calculated as:

P(x = 18) - P(x = 14)

0.92471 - 0.144

= 0.78071

c. If a cat lives to be over 18 years, would that be unusual? Why or why not?

For x > 18 years

z = 18 - 15.7/1.6

z= 1.4375

Probability value from Z-Table:

P(x<18) = 0.92471

P(x>18) = 1 - P(x<18) = 0.075288

Converting this to percentage:

0.075288 × 100 = 7.5288%

Hence, 7.5288% of the cats live to be over 18 years. Hence, it is not unusual.

d. How old would a cat have to be to be older than 90% of other cats?

From the question above, 10% of the cats would be older than 90% of other cats.

Hence, we find the z score of the 10th percentile

= -1.282

Hence,

-1.282 = x - 15.7/1.6

Cross Multiply

-1.282 × 1.6 = x - 15.7

- 2.0512 = x - 15.7

x = 15.7 -2.0512

x = 13.6488 years old

Approximately = 13.65 years old

3 0

3 years ago

After the lunch rush, Ellen had used up all the small containers and half the large containers. Which of these expressions can b

Sergeu [11.5K]

Answer:

$Remaining\ containers=\frac{y}{2}\\\\Where\ y\ is\ the\ number\ of\ large\ containers.$

Step-by-step explanation:

Let number of small containers $=x$

Let number of large containers $=y$

Total containers $=x+y$

<em>Used Containers:</em>

Small containers used $=x\ \ \ (as\ all\ small\ containers\ used)$

Larger containers used $=\frac{y}{2}\ \ \ \ \ \ (half\ of\ the\ large\ containers\ used)$

Total containers used $=x+\frac{y}{2}$

Remaining containers=total containers-used containers

$Remaining\ containers=(x+y)-(x+\frac{y}{2})=\frac{y}{2}\\\\\\\\If\ small\ containers(x)=30\\\\Large\ containers(y)=20\\\\total\ containers(x+y)=30\\\\Remaining\ containers=30-10-\frac{1}{2}\times 20$