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Nikolay [14]
3 years ago
11

Gabe makes a base salary of $1,500 per month He also earns a 3% commission on all of his sales What must the amount of his month

ly sales be for him to earn at least $3,000 per month? ​
Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0

Answer:

ɡabemakes a base salary of 1500 per month

he also earns a 3 commision onall of hos sales what must the amount of hid

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What is the size of angle c?
lesantik [10]

Answer:

55°

Step-by-step explanation:

d = 180 - 80 = 100

100 + 130 + 90 + (180 - 85) + (180 - c)

= 540

595 - c = 540

c = 595 - 540

c = 55°

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Solve for equation x. <br> 3In(x)+2In(4)=In(128)
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3 years ago
Witold says that 50% of a number will always be greater than 20% of any other number. Complete one inequality to support Witold'
Lelu [443]
.50 x 2 < .20 x 100
1 < 20

50% of 2 is not greater than 20% of 100.
4 0
3 years ago
I got 8 green crayons for the a part just confused for the b part
Shalnov [3]

Answer:

Just apply the same method to part B

x:24

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8 0
2 years ago
Read 2 more answers
When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis
Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so \mu = 22, \sigma = 4.

What is the probability that a fisherman catches a spotted seatrout within the legal limits?

They must be between 10 and 30 inches.

So, this is the pvalue of the Z score of X = 30 subtracted by the pvalue of the Z score of X = 10

X = 30

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

Z = \frac{30 - 22}{4}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 10

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

Z = \frac{10 - 22}{4}

Z = -3

Z = -3 has a pvalue of 0.00135.

This means that there is a 0.9772-0.00135 = 0.97585 = 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

3 0
2 years ago
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