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3 years ago

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Paul creates a scatter plot with a negative association . The x-axis of the scatter plot is titled “minutes spent at the mall” w

hich label is most likely the title of the y-axis of Paul’s scatter plot?
A. Distance walked
B. Money available to spend
C. Number of stores visited
D. Number of shirts purchased
( but I know the answer is b but I need to know why the other answers aren’t correct)

1 answer:

Y_Kistochka [10]3 years ago

4 0

Because as he spends time in the mall he has less money (he spends it) all the rest are positive associations

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A multiple-choice examination has 15 questions, each with five answers, only one of which is correct. Suppose that one of the st

Alex

Answer:

0.0111% probability that he answers at least 10 questions correctly

Step-by-step explanation:

For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.

This means that $n = 15, p = \frac{1}{5} = 0.2$

What is the probability that he answers at least 10 questions correctly?

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001$

$P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011$

$P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0$

$P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0$

$P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0$

$P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0$

$P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111$

0.0111% probability that he answers at least 10 questions correctly

3 0

2 years ago

please help. i’ve been stuck on this question the past 10 minutes i’ve tried multiple times and still don’t get it. i need help

Softa [21]

Answer:

To solve the equation, you have to to isolate the variable. So, first you subtract both sides by 7.

x +7 - 7 = -2 - 7

When you simplify further,

x = -9.

Hope this helps! :)

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2 years ago

Draw a circle and then draw the following parts on the diagram.

poizon [28]

Here's a photo of the diagram

3 0

3 years ago

In the figure, AB║CD and ∠EIA is congruent to ∠GJB Complete the following statements to prove that ∠IKL is congruent to ∠DLH.

REY [17]

Here it is given that AB || CD

< EIA = <GJB

Now

∠EIA ≅ ∠IKC and ∠GJB is ≅ ∠ JLD (Corresponding angles)

∠EIA ≅ ∠GJB then ∠IKC ≅ ∠ JLD (Substitution Property of Congruency)

∠IKL + ∠IKC 180° and ∠DLH + ∠JLD =180° (Linear Pair Theorem)

So

m∠IKL + m∠IKC = 180° ....(1)

But ∠IKC ≅ ∠JLD

m∠IKC = m∠JLD (SUBTRACTION PROPERTY OF CONGRUENCY)

So we have

m∠IKL + m∠JLD = 180°

∠IKL and ∠JLD are supplementary angles.

But ∠DLH and ∠JLD are supplementary angles.

∠IKL ≅ ∠DLH (CONGRUENT SUPPLEMENTS THEOREM)

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2 years ago

Read 2 more answers

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro

lisov135 [29]

Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that $\mu = 0, \sigma = 1$

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{-1.69 - 0}{1}$

$Z = -1.69$

$Z = -1.69$ has a p-value of 0.0455

X = -2.23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{-2.23 - 0}{1}$

$Z = -2.23$

$Z = -2.23$ has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch:

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3 years ago