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3 years ago

Find the value of x help pls

Mathematics

1 answer:

Dominik [7]3 years ago

6 0

Answer:

x = 29

Step-by-step explanation:

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10. Use substitution to solve the system of equations below.<br> - x + 2y = 5<br> 3x - 5y =-11

Liula [17]

Answer:

the answer for y is 4 and x is 3

y=4

x=3

6 0

3 years ago

PLEASE HELP I WILL MARK BRAINLIEST

lukranit [14]

82 plus 47 is 134 so angle 3 would be 46 degrees. So yes

5 0

3 years ago

Factor the polynomial completely<br> 10³ -25r² - 12r + 30

Olin [163]

ANSWER
(5r^2 - 6)(2r - 5)

EXPLANATION
Factor by grouping
(10r^3 - 25r^2) + (-12r + 30)

Factor a GCF
5r^2 (2r - 5) -6 (2r - 5)

Combine groups
(5r^2 - 6)(2r - 5)

CHECK
Use FOIL

First: 5r^2*2r = 10r^3
Outside: 5r^2*-5 = -25r^2
Inside: -6*2r = -12r
Last: -6*-5 = 30

5 0

2 years ago

You order seventeen burritos to go from a Mexican restaurant, seven with hot peppers and ten without. However, the restaurant fo

Kisachek [45]

Answer:

$P=0.369$

Step-by-step explanation:

In this binomial probability case, the success is to pick a burrito with hot peppers, there are 7 out of 17, the probability is p=7/17=0.4118. The formula is

$P=_nC_kp^k(1-p)^{n-k}$

Where n is the number of burritos taken at random, k is the number of success (burrito with hot peppers), p is the probability of success. $_nC_k$ denotes the number of combinations and its formula is:

$_nC_k=\frac{n!}{k!(n-k)!}$

At least two burritos have hot peppers means having two or three burritos, these two cases are calculated and sum to get the probability of at least two burritos that have hot peppers.

$P=_nC_kp^k(1-p)^{n-k}=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$

$P_2=\frac{3!}{2!(3-2)!}(0.4118)^2(1-0.4118)^{3-2}=\frac{3\times2!}{2!1!} (0.1696)(0.5882)^1\\P_2=3 (0.1696)(0.5882)=0.2993$

$P_2=\frac{3!}{3!(3-3)!}(0.4118)^3(1-0.4118)^{3-3}=\frac{1}{0!} (0.0698)(0.5882)^0\\P_2=1 (0.0698)(1)=0.0698$

$P=P_2+P_3=0.2993+0.0698=0.3691$

Round to three decimal places: P=0.369

7 0

4 years ago

A gumball machine contains 300 grape flavored balls, 400 cherry flavored balls, and 500 lemon flavored balls. What is the probab

Svet_ta [14]

Answer:

Pr(of getting 1grape ball, 1 cherry ball and 1 lemon ball) =0.20832

Step-by-step explanation:

Let the grape falvored ball be represented by G,

Cherry flavored balls be C

Lemon flavored balls be L

the possible order of picks for getting 1grape ball, 1 cherry ball and 1lemon ball are; GCL or GLC or CGL or CLG or LGC or LCG

Pr(of getting 1grape ball, 1 cherry ball and 1 lemon ball)

= Pr(GCL) or Pr(GLC) or Pr(CGL) or Pr(CLG) or Pr(LGC) or Pr(LCG)

with replacement we have;

Probability = $\frac{number of required outcomes}{number of possible outcomes}$

= $\frac{300}{1200}*\frac{400}{1200}*\frac{500}{1200} +\frac{300}{1200}*\frac{500}{1200}*\frac{400}{1200} +\frac{400}{1200}*\frac{300}{1200}*\frac{500}{1200} \\ \\+\frac{400}{1200}*\frac{500}{1200}*\frac{300}{1200} +\frac{500}{1200}*\frac{300}{1200}*\frac{400}{1200} +\frac{500}{1200}*\frac{400}{1200}*\frac{300}{1200}$

= 0.03472 + 0.03472 +0.03472 + 0.03472 + 0.03472 + 0.03472

=0.20832

3 0

3 years ago