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3 years ago

Which equation best represents the line graphed above? y= x+4 y=x+4 y=x-4 y = -x - 4

Mathematics

2 answers:

Brums [2.3K]3 years ago

6 0

Could you put a picture on?

Stolb23 [73]3 years ago

4 0

Y equals x plus four

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What value of q make the equation true 20/q 7/9

maria [59]

20/q = 7/9
20×9 = 7q
7q = 180
q = 180/7

5 0

2 years ago

Luke has $21 more than rachel and $48 more than daniel. altogether they have $168. how much does luke have?

Anettt [7]

this can be solve by 3 variable equation

let x be the money of luke

y money of rachel

z money of daniel

first equation

x = y + 21

second equation

x = z + 48

third equation

x + y + z = 168

solving simultaneously

x =79

y = 58

z = 31

7 0

3 years ago

21+77 using gcf and distributive property pls

defon

Step-by-step explanation:

u want to be friends ????????????

5 0

2 years ago

Read 2 more answers

At the movie theater Leon can buy 4 large popcorns or 2 large popcorns and 3 candy bars for the same price. If the candy bars co

AlexFokin [52]

4p = 2p + 3c....p = popcorn, c = candy bars
c = 6

4p = 2p + 3(6)
4p = 2p + 18
4p - 2p = 18
2p = 18
p = 18/2
p = 9......whoa....9 bucks for popcorn....nothing is cheap anymore

5 0

3 years ago

Question 4 of 10, Step 1 of 1 1/out of 10 Correct Certify Completion Icon Tries remaining:0 The Magazine Mass Marketing Company

erastovalidia [21]

Answer:

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

Step-by-step explanation:

For each entry form, there are only two possible outcomes. Either it includes an order, or it does not. The probability of an entry including an order is independent of any other entry, 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.

The Magazine Mass Marketing Company has received 16 entries in its latest sweepstakes.

This means that $n = 16$

They know that the probability of receiving a magazine subscription order with an entry form is 0.5.

This means that $p = 0.5$

What is the probability that no more than 3 of the entry forms will include an order?

At most 3 including an order, which is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

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

$P(X = 0) = C_{16,0}.(0.5)^{0}.(0.5)^{16} \approx 0$

$P(X = 1) = C_{16,1}.(0.5)^{1}.(0.5)^{15} = 0.0002$

$P(X = 2) = C_{16,2}.(0.5)^{2}.(0.5)^{14} = 0.0018$

$P(X = 3) = C_{16,3}.(0.5)^{3}.(0.5)^{13} = 0.0085$

Then

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0002 + 0.0018 + 0.0085 = 0.0105$

0.0105 = 1.05% probability that no more than 3 of the entry forms will include an order.

6 0

2 years ago