How far can your car go on one tank of gas if the tank holds 16 gallons and you average 23 miles of driving per gallon?

If f (x) = x2 – 2x, find:<br> f(6) = [?]

Alexxx [7]

Answer:

f(6) = 24

Step-by-step explanation:

f(x) = x^2 - 2x

Let x = 6

f(6) = 6^2 - 2(6)

= 36 - 12

= 24

6 0

2 years ago

Stephanie is 20 years old and has a base annual premium of 930 and a rating factor of 1.30. what is her total premium?

fredd [130]

Don't listen to the other guy. The ANSWER is A). 1209. Your welcome!

5 0

3 years ago

Last month, Katie started training for the swim team. She swam 20 laps every day for 18 training days. This month, Katie wants t

nasty-shy [4]

She swim 20 laps every day for 18 training days

so total laps done last month = 20*18 = 360 laps

She will swim 3 more laps per training day and add more training days to her monthly schedule.

20 laps + 3 laps = 23 laps

(a)If Katie adds 5 training days to her schedule this month, will she swim 215 more laps?

18 days + 5 days more = 23 days

23 laps every day for 23 training days

so total laps = 23*23 = 529 laps. That is 169 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

If Katie adds 10 more training days to her schedule this month, will she swim 215 more laps?

18 days + 10 days more = 28 days

23 laps every day for 28 training days

so total laps = 23*28 =644 laps. That is 284 more laps than 360 laps done last month. Hence she can swim 215 more laps

(c)If Katie adds 6 more training days to her schedule this month, will she swim 215 more laps?

18 days + 6 days more = 24 days

23 laps every day for 24 training days

so total laps = 23*24 =552 laps. That is 192 more laps than 360 laps done last month. Hence she cannot swim 215 more laps

Build an equation for finding the number of additional training days she needs to swim 215 more laps

Let x be the number of days more

So training days = 18 + x

215 more laps , so total laps = 360 + 215 = 575

3 more laps per training day so 20 + 3 = 23 laps per day

laps * training days = total laps

so equation becomes

23 * (18+x) = 575

Now solve for x

414 + 23x = 575

Subtract 414 from both sides

23x = 161

x= 7

so she needs 7 training days to swim 215 more laps

3 0

3 years ago

The height of a women in the united states are normally distributed with a mean of 64 inches and a standard deviation of 2.75 in

Tatiana [17]

Answer:

The area under the curve that represents the percent of women whose heights are at least 64 inches is 0.5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X, or the area under the curve representing values that are lower than x. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, which is the same as the area under the curve representing values that are higher than x.

In this problem, we have that:

$\mu = 64, \sigma = 2.75$