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ivolga24 [154]
3 years ago
14

How much do you get paid every two weeks if you work 20 hours a week.

Mathematics
2 answers:
Zarrin [17]3 years ago
6 0

Answer:

I get paid with insomia

Step-by-step explanation:

DaniilM [7]3 years ago
5 0

Answer:

328$

Step-by-step explanation:

20/weekly

20 × 2 = 40

40h/2 weeks

8.20 × 40 = 328$ for 2 weeks

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I mark brainliest! please help!<br>which graph shows all the values that satisfy 2/9x+ 3&gt; 4 5/9​
disa [49]

Answer:

The first option

Step-by-step explanation:

2/9 x + 3 > 4 5/9

2/9x > 4 5/9 - 3

2/9x > 14/9

x > 14/9 ÷ 2/9

x > 7

6 0
3 years ago
Read 2 more answers
The atmospheric pressure on an object decreases as altitude increases. If a is the height (in km) above sea level, then the pres
seraphim [82]

Answer:

373.8mmHg

Step-by-step explanation:

a =height (in km) above sea level,

the pressure P(a) (in mmHg) is approximated given as

P(a) = 760e–0.13a .

To determine the atmospheric pressure at 5.458 km, then we will input into the equation

P(5.458km) = 760e–0.13a .

= 760e^(-0.13×5.458)

=760e^-(0.70954)

= 760×0.4919

=373.8mmHg

Therefore, the atmospheric pressure at 5.458 km is 373.8mmHg

3 0
3 years ago
Please solve fast it's urgent​
Paul [167]

Step-by-step explanation:

hope it's helpful to you ☺️

4 0
3 years ago
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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
3 years ago
Is 1/5 and 3/10 eqivallant
kirill [66]

Answer:

No

1/5 is equivalent to 2/10

4 0
3 years ago
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