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ExtremeBDS [4]
3 years ago
13

Two whole numbers closest to 200

Mathematics
1 answer:
Temka [501]3 years ago
7 0
2 whole numbers is closer to 200 hundred is 199 n 198
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Is 15,36,39 a right triangle
NeTakaya

Answer:

Yes, is a right triangle

Step-by-step explanation:

To know if you can build a right triangle with those sides we have to make the following equality

h² = l1² + l2²

The hypotenuse is always the longest side so it should be 39

we check the equality of the equation

39² = 15² + 36²

1521 = 225 + 1296

1521 = 1521

Equality was fulfilled so it is a right triangle

7 0
2 years ago
What are the x– and y–intercepts of the graph of y = 2x^2 – 8x – 10?
Nadya [2.5K]
<u>x-intercepts are found by setting y=0</u>
y=2x^2-8x-10\rightarrow0=2x^2-8x-10
<em>factor out a 2</em> 0=2(x^2-4x-5)\rightarrow0=x^2-4x-5
<em />0=(x-5)(x+1) <em>roots should be </em>x=-1,5
therefore, x-intercepts are (-1,0) and (5,0)
<u>
</u><u>y-intercepts are found by setting x=0</u>
y=2x^2-8x-10\rightarrow y=2(0)^2-8(0)-10
\rightarrow y=0-0-10=-10
therefore, y-intercept is (0,-10)
4 0
3 years ago
Write four numbers that fall between 0.123 and 0.124
irga5000 [103]
That's what I think but if im wrong im sorry:
0.123-0.1231,0.1232, 0.1233, 0.1234-0.124
6 0
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
Solve mathematics problem in photo please! Second
Sholpan [36]
Hello,

8+h>2+3h
==>-2h>2-8
==>-2h>-6
==>h<3

Answer C

4 0
3 years ago
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