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3 years ago

The point (3,-1) is on the line given by which equation below? HELPP MEE PLSS

Mathematics

1 answer:

faust18 [17]3 years ago

7 0

Answer:

your answer will be <em><u>C. y = x-4</u></em>

Step-by-step explanation:

hope it helps you...

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Find the 52nd term of the given arithmetic sequence-30, -38, -46, -54...

torisob [31]

The sequence is given to be:

$-30,-38,-46,-54...$

The nth term of an arithmetic sequence is calculated using the formula:

$a_n=a_1+(n-1)d$

where a₁ is the first term, n is the number of terms, and d is the common difference.

From the given sequence, the following parameters can be gotten:

$\begin{gathered} a_1=-30 \\ d=-38-(-30)=-8 \end{gathered}$

Therefore, the parameters can be substituted into the formula to find the 52nd term:

$\begin{gathered} n=52 \\ \therefore \\ a_{52}=-30+(52-1)(-8) \\ a_{52}=-30+51(-8)=-30-408 \\ a_{52}=-438 \end{gathered}$

The 52nd term is -438.

8 0

1 year ago

I need help making the sentence make sense

Bingel [31]

Answer:

first line is mode second line is 21

4 0

3 years ago

Can anyone solve the rest pls

Juliette [100K]

X=-1 y=-1, x=1 y=5, x=3 y=11

5 0

3 years ago

Read 2 more answers

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

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

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

3 years ago

Jack wants to fill a rectangular box with sand. The length of the sand box is 3 feet, width is 6 inches, and height is 2.4 inche

Aleksandr-060686 [28]

Jack would need to buy 2 bags of sand.
To find the answer you first have to find the volume of the rectangular box which can be found using the formula length times width times height (remember that the numbers have to have the same unit so it would look like 3 times .5 times .2 if you convert them all the feet). the volume of the box is equal to .3
Then you could set the volume of the box equal to .15x to find how many boxes he would need. The formula would be .15x=.3
Divide by .15 to get x alone and you get 2 for the answer.

3 0

3 years ago

Read 2 more answers