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3 years ago

Hey can you please help me posted picture of question

Mathematics

1 answer:

RUDIKE [14]3 years ago

5 0

This is True.

If two events are independent the occurrence of one event is not affecting the occurrence of the other event. So if event B has already occurred it will not affect the probability of occurrence of event A.

Since P(A|B) represents probability of occurrence of event A, given that event B has already occurred, P(A|B) will be equal to P(A) i.e. the probability of occurrence of event A.

This means, no matters if event B has occurred or not, it will not have any impact on event A. Event A will occur irrespective of the event B.

So, the correct answer is TRUE.

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Use the graph to determine which statement describes f(x).

Lerok [7]

Answer: B

Step-by-step explanation:

7 0

2 years ago

PLZ Help Victoria used a probability simulator to pull 3 colored marbles from a bag and flip a coin 50 times. The results are sh

zmey [24]

the probability of getting tails up is 30/50 or most likely

the probability of getting a blue marble is 18/5 or least likely

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3 years ago

A glass paperweight has a composite shape: a square pyramid fitting exacty on top of an 8 centimeter cube. The pyramid has a hei

ArbitrLikvidat [17]

Answer:

Part 1) The volume of the paperweight is $576\ cm^{3}$

Part 2) The total surface area of the paperweight is $400\ cm^{2}$

Step-by-step explanation:

Part 1) what is the volume of the paperweight?

we know that

The volume of the paperweight is equal to the volume of the square pyramid plus the volume of the cube

step 1

Find the volume of the pyramid

The volume of the pyramid is equal to

$V=\frac{1}{3}BH$

where

B is the area of the square base

H is the height of the pyramid

$B=8^{2}=64\ cm^{2}$

$H=3\ cm$

substitute

$V=\frac{1}{3}(64)(3)=64\ cm^{3}$

step 2

Find the volume of the cube

The volume of the cube is equal to

$V=b^{3}$

$V=8^{3}=512\ cm^{3}$

step 3

Find the volume of the paperweight

$64\ cm^{3}+512\ cm^{3}=576\ cm^{3}$

Part 2) what is the total surface area of the paperweight?

we know that

The total surface area of the paperweight is equal to the surface area of 5 faces of the cube plus the lateral area of the pyramid

step 1

Find the surface area of 5 faces of the cube

$SA=5b^{2}$

$SA=5(8^{2})=320\ cm^{2}$

step 2

Find the lateral area of the pyramid

$LA=4[\frac{1}{2}bh]$

$LA=4[\frac{1}{2}(8)(5)]=80\ cm^{2}$

step 3

Find the total surface area of the paperweight

$320\ cm^{2}+80\ cm^{2}=400\ cm^{2}$

8 0

3 years ago

Consider the set T = {1, 2, 3, 4, 5, 6}. Let R = {(a, b) | a divides b} be a relation on the set T. Identify the missing ordered

serg [7]

Answer: The missing ordered pairs are (2, 2), (2, 4), (2, 6), (3, 3) and (3, 6).

Step-by-step explanation: We are given the following set :

T = {1, 2, 3, 4, 5, 6}.

And, R = {(a, b) | a divides b} be a relation on the set T.

We are to identify the missing pairs from the following list of all ordered pairs in the given relation R on the set T :

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), _____, (4, 4), (5, 5), and (6, 6).}.

Since R contains the ordered pairs (a, b) such that a divides b, so all the ordered pairs in R are

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (5, 5), and (6, 6).

Thus, the missing ordered pairs are (2, 2), (2, 4), (2, 6), (3, 3) and (3, 6).

4 0

3 years ago

According to the National Vital Statistics, full-term babies' birth weights are Normally distributed with a mean of 7.5 pounds a

Sav [38]

Answer:

68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 7.5, \sigma = 1.1$