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3 years ago

Dakota tiled a floor with tiles that look like this:

Mathematics

1 answer:

Sliva [168]3 years ago

3 0

AB=8 in.
AC=6 in.
BC=10 in.
<ABC=37
<CAB=90
<ACB=53
<BCE=2*37=74
<CED=2*53=106
l=6*AB=6*8=48 in.
w=6*AC=6*6=36 in.

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Help me plss to find the area

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Answer:

A=88 square units

Step-by-step explanation:

A=B x H

A= 11 x 8

A=88 square units

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3 years ago

A bag contains 60 pieces of candy with different flavors: grape, strawberry, lemon, orange, apple, and watermelon. You want to r

a_sh-v [17]

Answer:

We would use to describe numbers from 1 - 6.

Step-by-step explanation:

We represent the flavors as a set in Roster form.

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podryga [215]

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Step-by-step explanation:

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We learned in that about 69.7% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty

maw [93]

Answer:

(1) The expected number of people who would have consumed alcoholic beverages is 34.9.

(2) The standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3) It is surprising that there were 45 or more people who have consumed alcoholic beverages.

Step-by-step explanation:

Let X = number of adults between 18 to 20 years consumed alcoholic beverages in 2008.

The probability of the random variable X is, p = 0.697.

A random sample of n = 50 adults in the age group 18 - 20 years is selected.

An adult, in the age group 18 - 20 years, consuming alcohol is independent of the others.

The random variable X follows a Binomial distribution with parameters n = 50 and p = 0.697.

The probability mass function of a Binomial random variable X is:

$P(X=x)={50\choose x}0.697^{x}(1-0.697)^{50-x};\ x=0,1,2,3...$

(1)

Compute the expected value of X as follows:

$E(X)=np\\=50\times 0.697\\=34.85\\\approx34.9$

Thus, the expected number of people who would have consumed alcoholic beverages is 34.9.

(2)

Compute the standard deviation of X as follows:

$SD(X)=\sqrt{np(1-p)}=\sqrt{50\times 0.697\times (1-0.697)}=10.55955\approx10.56$

Thus, the standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3)

Compute the probability of X ≥ 45 as follows:

P (X ≥ 45) = P (X = 45) + P (X = 46) + ... + P (X = 50)

$=\sum\limits^{50}_{x=45} {50\choose x}0.697^{x}(1-0.697)^{50-x}\\=0.0005+0.0001+0.00002+0.000003+0+0\\=0.000623\\\approx0.0006$

The probability that 45 or more have consumed alcoholic beverages is 0.0006.

An unusual or surprising event is an event that has a very low probability of success, i.e. p < 0.05.

The probability of 45 or more have consumed alcoholic beverages is 0.0006. This probability value is very small.

Thus, it is surprising that there were 45 or more people who have consumed alcoholic beverages.

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2 years ago