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2 years ago

5 less the quotient of 14 times a number x and 9

Mathematics

1 answer:

AURORKA [14]2 years ago

7 0

Answer:

$\frac{14x}{9}-5$

Step-by-step explanation:

14 times a number x : 14* x = 14x

Quotient of 14 times a number x an 9 : $\frac{14x}{9}$

5 less the quotient of 14 times a number x and 9 :

$\frac{14x}{9}-5$

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Starting at home, Luis traveled uphill to the gift store for 50 minutes at just 6 mph. He then traveled back home along the same

mario62 [17]

Average speed for the entire trip, both ways, is

   (Total distance) divided by (total time) .

We don't know the distance from his house to the gift store,
and we don't know how long it took him to get back.
We'll need to calculate these.

-- On the trip TO the store, it took him 50 minutes, at 6 mph.
-- 50 minutes is 5/6 of an hour.
-- Traveling at 6 mph for 5/6 of an hour, he covered 5 miles.
-- The gift store is 5 miles from his house.
-- The total trip both ways was 10 miles.

-- On the way BACK home from the store, he moved at 12 mph.
-- Going 5 miles at 12 mph, it takes (5/12 hour) = 25 minutes.

Now we have everything we need.

Distance:
   Going:    5 miles
   Returning: 5 miles
   Total    10 miles

Time:
   Going:    50 minutes
   Returning:   25 minutes
   Total:    75 minutes = 1.25 hours

   Average speed for the whole trip =

   (total distance) / (total time)

   = (10 miles) / (1.25 hours)

   = (10 / 1.25) miles/hours

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3 years ago

Geometry draw a quadrilateral with the verteices r(-3,-1), s(-1,0),t

agasfer [191]

This is an incomplete problem

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3 years ago

A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi

34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{X}=n\mu$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{X}=\sqrt{n}\sigma$

It is provided that:

$\mu=\$27\\\sigma=\$18\\n=310$

As the sample size is quite large, i.e. n = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

$P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z$

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let s denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding z-value is, -2.33.

Compute the value of s as follows:

$z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57$

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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vesna_86 [32]

Answer:

Step-by-step explanation:

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