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2 years ago

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On discovering that her family had a 70% risk of heart attack, Erin took a treadmill test to check her own potential of having a

heart attack. The doctors told her that the reliability of the stress test is 67%. What is the probability that Erin will have a heart attack and the test predicts it?
0.099

0.201

0.231

0.469

2 answers:

marshall27 [118]2 years ago

6 0

Answer:

Hence, the probability that Erin will have a heart attack and the test predicts it is:

0.469

Step-by-step explanation:

On discovering that her family had a 70% risk of heart attack, Erin took a treadmill test to check her own potential of having a heart attack.

The doctors told her that the reliability of the stress test is 67%.

The probability that she will have an heart attack and the test predicts it is:

70% of 67%.

$=\dfrac{70}{100}\times \dfrac{67}{100}\\\\=0.469$

Hence, the probability that Erin will have a heart attack and the test predicts it is:

0.469

ikadub [295]2 years ago

5 0

Answer:on edmentum d is wrong but idk wats right answer

Step-by-step explanation: took test and got it wrong

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The average life of a bread-making machine is 7 years, with a standard deviation of 1 year. Assuming that the lives of these mac

Alina [70]

Answer:

a) $P(6.4$

b) $a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable life of a bread making machine. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =7,\sigma =1)$

We take a sample of n=9 . That represent the sample size.

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=7, \frac{1}{\sqrt{9}})$

Solution to the problem

Part a

(a) the probability that the mean life of a random sample of 9 such machines falls between 6.4 and 7.2

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{1}{\sqrt{9}}=0.333$

We want this probability:

$P(6.4$

Part b

b) The value of x to the right of which 15% of the means computed from random samples of size 9 would fall.

For this part we want to find a value a, such that we satisfy this condition:

$P(\bar X>a)=0.15$ (a)

$P(\bar X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.036. On this case P(Z<1.036)=0.85 and P(Z>1.036)=0.15

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.036$

And if we solve for a we got

$a=7 +1.036*0.333=7.345$

So the value of bread-making machine that separates the bottom 85% of data from the top 15% is 7.345.

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Nick starts with 20 milligrams of a radioactive substance. The amount of the substance decreases by 12 each week for a number of

Gre4nikov [31]

Answer:

what the answer

Step-by-step explanation:

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As a computer technician, Andre makes $20 per hour to diagnose a problem and $25 per hour to fix a problem. He works fewer than

irinina [24]

For
ax+by=c
the slope is -a/b
20x+25y≥200
slope=-20/25=-4/5
negative slope

yint is where x=0
20(0)+25y≥200
25y≥200
y≥98
positive yint

x+y<10
slope=-1/1=-1
yint is where x=0
y<10
yint is at y=10

since it is equal, it is solid line
to tell if it is above then sub (0,0) and see if true
0≥200
false
shade on side that doesn't have (0,0), shade above line

x+y<10 doesn't have equal under so it is dashed
test (0,0)
0<10
true, it is shaded below

test point (4,5)

20(4)+25(5)≥200
80+125≥200
225≥200
true

4+5<10
9<10
true

so the ones that are true are

The line x + y < 10 has a negative slope and a positive y-intercept.
The line representing 20x + 25y ≥ 200 is solid and the graph is shaded above the line.
The overlapping region contains the point (4, 5).

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2 years ago

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Addie walked 2 1 2 miles in 45 minutes. Suzie covered 2 2 5 miles in 2 3 of an hour. Which girl walked faster? By how much?

kobusy [5.1K]

Answer:

Suzie walks faster by 0.27 miles per hour.

Step-by-step explanation:

Addie walked $2\frac{1}{2}$ miles in 45 minutes and Suzie covered $2\frac{2}{5}$ miles in $\frac{2}{3}$ of an hour.

So, Addie walked 2.5 miles in 45 minutes i.e. $\frac{45}{60} = 0.75$ hours.

Therefore, the walking speed of Addie is $\frac{2.5}{0.75} = 3.33$ miles per hour.

Again, Suzie covered $2\frac{2}{5}$ i.e. 2.4 miles in $\frac{2}{3}$ i.e. 0.67 hours.

So, the walking speed of Suzie is $\frac{2.4}{0.67} = 3.6$ miles per hour.

Hence, Suzie walks faster by (3.6 - 3.33) = 0.27 miles per hour. (Answer)

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2 years ago

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RUDIKE [14]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And for this case the 95% confidence interval is given by (2.13; 2.37)

We have a point of estimate for the sample mean with this formula:

$\bar X = \frac{Upper+ Lower}{2}= \frac{3.37+2.13}{2}= 2.75$

And for the margin of error we have the following estimation:

$ME= \frac{Upper -Lower}{2}= \frac{3.37-2.13}{2}= 0.62$

5 0

2 years ago