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3 years ago

The half-life of a radioactive substance is 11 days. After 12 days a sample of the substance has been reduced

Mathematics

1 answer:

sergiy2304 [10]3 years ago

8 0

The mass is 48.3 that’s it

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A student is attempting to solve the equation below for the variable x. Which of the statements below best applies to the mathem

Elanso [62]

Please answer please please thank you

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3 years ago

1/2(4x+10)=2 i need to get x

kari74 [83]

First you need to distribute the 1/2 by the numbers in the parentheses
1/2•4x and 1/2•10 which equals
2x+5=10

next, you will need to get the 2x by itself so you would do
2x+5=10
-5=-5
2x=5

finally, you divide the 2 by the 5 so you can get the x by itself
2x divided by 5

the answer is: x=2.5

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3 years ago

X and y last one <br><br><br>i put all the points i could on this question<br>

Rudik [331]

Step-by-step explanation:

15^2-12^2

225-144

√81=9

21÷9=2.3333(recurring)

2.33333×15=x

X=34.999

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3 years ago

Will give Brainly

g100num [7]

Answer: 33.3%

s Step-by-step explanation: The combined arc length of the red area is 120, which is 1/3 of 360, the full arc. So the probability is 1/3, or 33.3%. Hope this helps!

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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.

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3 years ago