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dangina [55]
3 years ago
12

What is the domain and range for this problem?

Mathematics
1 answer:
padilas [110]3 years ago
6 0

Answer:

domain: 2≤x≤12

range: -2≤x≤4

Step-by-step explanation:

x is range and y is domain

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Out of 5000 policies for cell phones, a company received 585 claims. The average amount paid on the claims was $80. If the annua
Ede4ka [16]

Answer:

the customer's expected value of the cellphone insurance policy is -$38.64. (Option C)

Step-by-step explanation:

Based on the data given, 585/5000 or 11.7% of the total number policies have claims. This means, 100% - 11.7% = 88.3% have not claimed their cellphone insurance.

  • In the  chart, we are focusing on the customer's end because the question is asking for the customer's expected value↓
  • In the chart, if the customer has no claims, then he losses his $48. However, if he has claims, he gains $32 since $80 - 48 = $32. Also, based on the data given by the company, 88.30% have no claims and only 11.7% of the customers have claims.
  • To get the expected value, we will multiply -48 and 88.3% as well as 32 and 11.7%. 48 is negative because it indicates a loss.

Therefore, your Answer is -38.64

6 0
3 years ago
How many solutions does y=-5x-1 y=-5x+7
weeeeeb [17]

Answer:

it has no solutions

Step-by-step explanation:

y = mx + b

y = -5x - 1......slope here is -5 and y int is -1

y = -5x + 7....slope here is -5 and y int is 7

5 0
3 years ago
The angle 01 is located in Quadrant IV, and sin(01) = -24/ 25<br><br> What is the value of cos(01)?​
pav-90 [236]

Answer:

\frac{7}{25}

Step-by-step explanation:

We can solve that using this identity,

\sin {}^{2} (x)  +  \cos {}^{2} (x)  = 1

plug in -24/25 from x in sin

\sin {}^{2} ( \frac{ - 24}{25} )  +  \cos {}^{2} (x)  = 1

\sin( \frac{576}{625} )  +  \cos {}^{2} (x)  = 1

\cos {}^{2} (x)  =  1 -  \frac{576}{625}

\cos {}^{2} (x)  =  \frac{49}{625}

\cos(x)  =  \frac{7}{25}

Since cos is positve in quadrant 4, the answer is 7/25.

4 0
3 years ago
Read 2 more answers
A marine biologist is setting up a net 6 feet below the surface of the water. each foot of the water screens out 40% of the ligh
just olya [345]

The approximate percentage of light remaining after passing through 6 feet of water, to the nearest tenth of a percent is 4.7%

<h3>How is the percentage of light remaining at a depth of 6 feet calculated?</h3>

The percentage of light remaining is calculated from the percentage of light remaining after a depth of one foot each.

Percentage of light remaining after 1 feet = 100 - 40% = 60%.

Percentage of light remaining after 6 feet = 60% × 60% × 60% × 60% × 60% × 60% = 4.7%

Therefore, the approximate percentage of light remaining after passing through 6 feet of water, to the nearest tenth of a percent is 4.7%.

Learn more about percentage at: brainly.com/question/24304697

#SPJ4

4 0
2 years ago
I also need a answer for these two
LUCKY_DIMON [66]

Answer: See explanation

Step-by-step explanation:

a. 19 > x + 4

Collect like terms

= 19 - 4 > x

= 15 > x

= x < 15

b. 5x > 4x - 9

Collect like terms

5x - 4x> -9

x > -9

7 0
3 years ago
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