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Ganezh [65]
2 years ago
8

I need help on this question please help me

Mathematics
1 answer:
Anastasy [175]2 years ago
6 0

Answer:

Step-by-step explanation:

6*[4-(42-63)]=

6*[4-(-21)]=

6*[4+21]=

6*25=150.

6*[4-9÷3]=

6*[4-27]=

6*(-23)=-138.

6*[4-3]=

6*1=6.

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The answer for this problem is c bc that’s what I think
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Find the value of x in the isosceles triangle shown below.​
Ghella [55]

Answer: A, x = square root of 60.

Step-by-step explanation:

Use the Pythagorean Theorem, a^2 + b^2 = c^2.

8^2 = 2^2 + x^2

x^2 = 8^2 - 2^2

x^2 = 64 - 4

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This figure has two intersecting lines and a ray. What is the value of x?
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The value of x is D. 37°

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3 years ago
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
Rasek [7]

Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}

\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

t_{11,0.025} = 2.201

The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0
3 years ago
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