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77julia77 [94]
3 years ago
15

3. You have spent a total of $440 total on your gym membership so far. If the gym charges a

Mathematics
1 answer:
slava [35]3 years ago
5 0

Answer:

13 months

Step-by-step explanation:

m=number of months

30(m)+50

30*13= 390

390+50=440

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What is the answer to this equation? (picture attached)
SashulF [63]

Answer:

x = \frac{42}{5}

Step-by-step explanation:

Given

19 - \frac{5x+6}{4} = 7 ( subtract 19 from both sides )

- \frac{5x+6}{4} = - 12

Multiply both sides by - 4 to clear the fraction and the negative signs

5x + 6 = 48 ( subtract 6 from both sides )

5x = 42 ( divide both sides by 5 )

x = \frac{42}{5}

6 0
3 years ago
Please help me with 19
andrew-mc [135]
The answer to 19th 2/3
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3 years ago
Read 2 more answers
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. <em>n</em> = 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 <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> 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.

5 0
2 years ago
The cost of a car rental is twenty-five dollars a day plus ten cents a mile. What is the total cost, in dollars, if the car is r
Fudgin [204]

Answer:

The total cost is D) $141

Step-by-step explanation:

To solve we can use the equation: y=25d+0.10m

d= number of days  m= number of miles

We know both of these values so we can plug them in and solve.

y= 25(3) + 0.10(660)

y= 75 + 66

y = 141

4 0
3 years ago
Jevon arrives home from school at 3:10.it takes him 15 minutes to eat a snack 1/2 hour to complete his homework 7 minutes to cha
zysi [14]
Arrives at home 3:10pm
snack 15min
homework 30min
change/pack 7min
bike ride 18min
arrives to soccer practice 4:20pm
6 0
3 years ago
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