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Grace [21]
3 years ago
6

Help ^^^^^^^^^^^^^^^^

Mathematics
1 answer:
il63 [147K]3 years ago
8 0
-18-12=-30 the answer is -30
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If the Semi Monthly is $4,000
Andreyy89
Disclaimer- this is all assuming that "semi monthly" means twice a month. if it doesn't, ignore this answer

A- if semi monthly is half a month, then 4 times 2 is 8, and 8 times 12 is 96,000.

B- If there are 4 weeks in a month and 4000$ accounts for half of that pay, then the weekly pay is 2000, making the bi-weekly pay 1000.

C- The monthly pay is 8,000$ because 4,000 x 2 is 8,000

D- The weekly pay is 2,000$ because the monthly pay is 8,000$ and there are 4 weeks in a month

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3 years ago
What is the equation?​
elena-s [515]

Answer:

  cost = 1.35n

  $33.75 for 25 songs

Step-by-step explanation:

Based on the numbers given, the cost is proportional to the number of songs downloaded. The constant of proportionality is the cost of one song: 1.35.

  cost = 1.35n

For 25 songs, ...

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The equation is ...

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5 0
3 years ago
Plzzzz help me asap WILL GIVE BRAINLIEST
Brrunno [24]
The answer will be d i believe
8 0
3 years ago
Read 2 more answers
The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0
3 years ago
When constructing an inscribed square how many lines will be drawn
Oksanka [162]
From the research I just did, it seems that 6 lines will be drawn when constructing an inscribed square.

Hope this helped :)
4 0
3 years ago
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