Idk what to do, can someone please help?

slader Suppose that 8% of all bicycle racers use steroids, that a bicyclist who uses steroids tests positive for steroids 96% of

d1i1m1o1n [39]

Answer:

P = 0.4812

Step-by-step explanation:

First, we need to use here two expressions and then do the calculations.

The first one is the conditional probability which is:

P(B|A) = P(A∩B)/P(A) (1)

The second expression to use has relation with the Bayes's theorem which is the following:

P(D|C) = P(C|D)*P(D) / P(C|D)*P(D) + P(C|d)*P(d) (2)

Now, the expression (2) is the one that we will use to calculate the probability of a selected random bicyclist who tests positive for steroids.

So, in this case, we will call C for positive and D that is using steroids and d is the opposite of d, which means do not use steroids.

Then, the probabilities are the following:

P(D) = 8% or 0.08

P(C|D) = 96% or 0.96

P(C|d) = 9% or 0.09

P(d) = 1 - 0.08 = 0.92

With these data, let's replace in expression 2

P(D|C) = 0.96 * 0.08 /0.96 * 0.08 + 0.09*0.92

P(D|C) = 0.0768 / 0.1596

P(D|C) = 0.4812 or 48.12%

8 0

2 years ago

Manuel sold 200liters of soda at a baseball game. How much is this in milliliters?Be sure to include the correct unit in your an

Sever21 [200]

1 liter is equal to 1000 milliliters, so we can multiply 200 liters by 1000 to get 200000 milliliters, which should be your answer. In general, the "milli-" prefix denotes a factor of one thousand times smaller, so 1 meter is equal to 1000 millimeters, 1 second is equal to 1000 milliseconds, and so on.

5 0

2 years ago

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones

Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal( $\mu=237,\sigma^{2} =54^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean family spending = $237

$\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

= 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

= 1 - 0.8749 = 0.1251

P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

= 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101

5 0

2 years ago

HELP!

loris [4]

<h3>Answer: Choice C, Ro, 270 </h3>

270 degree counterclockwise rotation around the origin

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Check out the diagram below. In the diagram, I only focus on points A and A'.

We have A = (-3,4) move to A ' = (4, 3). We see the x and y coordinates swap places. Then the new second coordinate -3 becomes positive, or it has flipped sign. So the rule applied here is $(x,y) \to (y,-x)$ which describes a 90 degree clockwise rotation; which is equivalent to a 270 degree counterclockwise rotation