If DE is congruent to KR and O is the midpoint of ER and DK, which of the following congruence postulates can be used

A library is buying books at a constant rate of 150 books per year. The equation

zloy xaker [14]

Answer:

C

Step-by-step explanation:

The answer is C, because X is per year, so its 150 books per year, and the library already had is 7000 bc your adding that number to the amount of books your going to buy per year

7 0

3 years ago

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The teen club had a cupcake sale. they started with 126 cupcakes. they sold 2/3 of their cupcakes in the morning. how many were

Readme [11.4K]

126 ÷ 3= 42
42×2= 84 (number of cupcakes sold)
126-84= 42 (number of cupcakes left)

3 0

3 years ago

Is this an expression or an equation?<br> 32 + 8(4)<br> expression<br> equation

KonstantinChe [14]

It is an expression, as an equation has an equal sign on it.

This has no equality property, therefore it is an expression.

6 0

3 years ago

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In psychology, there is a particular Mental Development Index (MDI) used in the study of infants. The scores on the MDI have app

ivann1987 [24]

Answer:

0.1587

Step-by-step explanation:

Here, mean=μ=100 and standard deviation=σ=16.

We have to find P(average MDI scores of 64 children > 102)=P(xbar>102).

n=64

μxbar=μ=100

σxbar=σ/√n=16/√64=16/8=2

P(xbar>102)=P((xbar-μxbar)/σxbar>(102-100)/2)

P(xbar>102)=P(z>1)

P(xbar>102)=P(0<z<∞)-P(0<z<1)

P(xbar>102)=0.5-0.3413

P(xbar>102)=0.1587

Thus, the probability that the average is greater than 102 is 15.87%

4 0

4 years ago

A professor knows that her statistics students' final exam scores have a mean of 79 and a standard deviation of 11.3. In his cla

OverLord2011 [107]

For each student, there are only two possible outcomes. Either they score an A, or they do not. The probability of a student scoring an A is independent of any other student, which means that the binomial probability distribution is used to solve this question.

Additionally, to find the proportion of students who scored an A, the normal distribution is used.

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 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.

Normal Probability Distribution

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

In a set with mean and standard deviation , 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.

Proportion of students that scored an A:

Scores have a mean of 79 and a standard deviation of 11.3, which means that $\mu = 79, \sigma = 11.3$

Scores of 90 or higher are graded an A, which means that the proportion is 1 subtracted by the p-value of Z when X = 90, so:

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

$Z = \frac{90 - 79}{11.3}$

$Z = 0.97$

$Z = 0.97$ has a p-value of 0.8340.

1 - 0.8340 = 0.166

The proportion of students that scored an A is 0.166.

Probability that 6 students or more will score an "A" on the final exam:

Binomial distribution.

22 students, which means that $n = 22$

The proportion of students that scored an A is 0.166, which means that $p = 0.166$

The probability is:

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Then

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

$P(X = 0) = C_{22,0}.(0.166)^{0}.(0.834)^{22} = 0.0184$

$P(X = 1) = C_{22,1}.(0.166)^{1}.(0.834)^{21} = 0.0807$

$P(X = 2) = C_{22,2}.(0.166)^{2}.(0.834)^{20} = 0.1687$

$P(X = 3) = C_{22,3}.(0.166)^{3}.(0.834)^{19} = 0.2239$

$P(X = 4) = C_{22,4}.(0.166)^{4}.(0.834)^{18} = 0.2117$

$P(X = 5) = C_{22,5}.(0.166)^{5}.(0.834)^{17} = 0.1517$

Then

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0184 + 0.0807 + 0.1687 + 0.2239 + 0.2117 + 0.1517 = 0.8551$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8551 = 0.1449$

Thus

0.1449 = 14.49% probability that 6 students or more will score an "A" on the final exam.

For a problem that used the normal distribution, you can check brainly.com/question/15181104, and for a problem that used the binomial distribution, you can check brainly.com/question/15557838

3 0

3 years ago