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3 years ago

PLEASE HELP ME IM TAKING TIMED TEST I WILL MARK YOU AS BRAINLEST!!

Mathematics

2 answers:

stepladder [879]3 years ago

7 0

Answer:

-4.5,-0.5,0.3,0.7,2.3

Step-by-step expla

tresset_1 [31]3 years ago

4 0

Answer:-4.5, -0.5, 0.3, 0.7, 2.3

Step-by-step explanation:

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Which angles are adjacent to each other? Question is located in the attached file (image)

daser333 [38]

Answer:

∠ 1 and ∠2

Step-by-step explanation:

Adjacent angles share a common ray or side between them, and have the same vertex.

Therefore, the adjacent angles from the given image are: ∠ 1 and ∠2.

The other given options are not adjacent angles because they do not have the same common vertex and ray.

8 0

2 years ago

What would you have to do to change 10 cubic feet into cubic inches?

Alina [70]

A. Divide by 46,656

B. Multiply by 1,728

C. Divide by 1,728

D. Multiply by 46,656

THE ANSWER IS B MULTIPLY BY 1,728

5 0

3 years ago

Trina planted plants in her garden last week. She planted twice as many cucumbers as zucchini and eight more tomatoes than cucum

maxonik [38]

Answer:

7 Zuchini, 14 Cucumbers and 22 Tomatoes plants.

Step-by-step explanation:

Let the number of tomatoes=t
Let the number of cucumber planted=c
Let the number of zucchini planted=z
Total Number of Plants=43

She planted twice as many cucumbers as zucchini, this is written as:

c=2z

She also planted eight more tomatoes than cucumbers, this is written as:

t=c+8=2z+8

In total, she planted 43 plants

Therefore:

z+(2z)+(2z+8)=43

5z+8=43

5z=43-8

5z=35

Divide both sides by 5

z=7

Number of Cucumbers, c=2z=2 X 7 =14

Number of Tomatoes, t=c+8=14+8=22

Therefore, she planted 7 Zuchini, 14 Cucumbers and 22 Tomatoes plants.

3 0

3 years ago

Math question for algebra 1! Please help thanks picture provided

Murrr4er [49]

The answer is 1800 you take .06 being the cost of every sq yard and multiply it by 300 being the amount of sq yards

5 0

3 years ago

According to a study, 50 % of adult smokers started smoking before 21 years old. 5 smokers 21 years old or older are randomly se

belka [17]

Answer:

a) The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

Step-by-step explanation:

For each smoker, there are only two possible outcomes. Either they started smoking before 21 years old, or they did not. Smokers are independent, which means that the binomial probability distribution is used 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.

50% of adult smokers started smoking before 21 years old.

This means that $p = 0.5$

5 smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

This means that $n = 5$ .

a) The probability that at least 2 of them started smoking before 21 years of age is

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{5,0}.(0.5)^{0}.(0.5)^{5} = 0.03125$

$P(X = 1) = C_{5,1}.(0.5)^{1}.(0.5)^{4} = 0.15625$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.03125 + 0.15625 = 0.1875$

The probability that at least 2 of them started smoking before 21 years of age is 0.1875 = 18.75%.

b) The probability that at most 4 of them started smoking before 21 years of age is

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

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

$P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.03125$

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.03125 = 0.96875$

The probability that at most 4 of them started smoking before 21 years of age is 0.96875 = 96.875%.

c) The probability that exactly 3 of them started smoking before 21 years of age is

This is P(X = 3). So

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

$P(X = 3) = C_{5,3}.(0.5)^{3}.(0.5)^{2} = 0.3125$

The probability that exactly 3 of them started smoking before 21 years of age is 0.3125 = 31.25%.

4 0

2 years ago