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inessss [21]
2 years ago
15

———- pls help!! ——-

Mathematics
2 answers:
nalin [4]2 years ago
6 0

According to above graph :

  • g(-4) = 6

so, correct option is C. 6

aleksley [76]2 years ago
6 0
Answer
C
Explanation
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Four siblings have a dog walking business. The table shows the hours worked by each sibling. Each sibling earns $25.50 per hour
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Step-by-step explanation:

Michael works for 25 hours

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the answer is in the picture

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3 years ago
In a sample of eight whitefish caught in Yellowknife Bay, the mean arsenic concentration in the liver was 0.32 mg/kg, with a sta
Advocard [28]

Answer:

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 7

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.3246

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.3646\frac{0.06}{\sqrt{8}} = 0.0502

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg

The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg

The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).

8 0
2 years ago
Using 50 observations, the following regression output is obtained from estimating y = β0 + β1x + β2d1 + β3d2 + ε. Coefficients
pickupchik [31]

Answer:

The answer is "789.03 and 806.16".

Step-by-step explanation:

y\^ \ \ = -0.45 + 3.78(x)  -11.88(d_1) + 5.25(d_2)

For point a:

Calculating the value for y\^  \ \  x = 212,\ \  d_1 = 1, and  \ \ d_2 = 0

y\^  = -0.45 + 3.78(212)  -11.88(1) + 5.25(0)

     = -0.45 + 801.36 -11.88 + 0\\\\= 789.03  

For point b:

Calculating the value of y\^ \ \ for \ \ x = 212, \ \ d_1 = 0, and \ \ d_2 = 1

y\^  = -0.45 + 3.78(212)  -11.88(0) + 5.25(1)  

    = -0.45 + 801.36 -0 + 5.25\\\\= 806.16

3 0
3 years ago
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