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WINSTONCH [101]
3 years ago
6

Drusilla replaces the light bulb in the hall closet every 9 months and replaces the air filter every 6 months will she replace b

oth the light bulb and the air filter?
Mathematics
1 answer:
Mazyrski [523]3 years ago
5 0
If she replaces the air filter every 6 months the light bulb will not be changed because there are still 3 months remaining. Therefore, no she will not replace both of them.
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Please answer the following 4 questions
Lilit [14]
1. The correct answer should be A
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4. The answer should be D


Hope this helps :)
7 0
3 years ago
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Which graph correctly represents (5/2)x - y < 3?
Setler79 [48]

You can rewrite the inequality as

y > \cfrac{5}{2}\ x - 3

Note that

y = f(x) = \cfrac{5}{2}\ x - 3

is the equation of the line drawn in the pictures. This means that, for every point on the line, the y coordinate is exactly image of the x coordinate, i.e. f(x)

The inequality is satisfied by all points whose y coordinate exceeds the image of the x coordinate. Since the y axis is positively oriented upwards, a greater value for the y coordinate means that the point has to be higher than the line.

Also, since the equality inlves a > and not a \geq sign, the line itself is excluded.

So, the correct graph is the second one.

3 0
3 years ago
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5<br> Find x to the nearest tenth.<br> 18.7<br> 13<br> 19.3<br> NEXT QUESTION<br> © ASK FOR HELP
Iteru [2.4K]
Think it’s option 1.3
6 0
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. <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
3 years ago
(2, -7) and (4, -13)
kkurt [141]

Answer:

-14 or (6,20)

8 0
2 years ago
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