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4 years ago

Solve the equation 3x + 4y = 16 for x.

Mathematics

1 answer:

Lisa [10]4 years ago

6 0

3x-4y=16

add 4y to both sides

3x=4y+16

divide both sides by 3

x=(4y+16)/3

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Need help asap pls!!!

lisabon 2012 [21]

Answer:

Step-by-step explanation:

According to the given description, PS is the mid segment of the $\triangle TQR$

Therefore by Mid-segment Theorem:

$PS = \frac{1}{2} QR \\ \\ PS = \frac{1}{2} \times 32 \\ \\ PS =16 \:$

3 0

3 years ago

Out of 450 applicants for a job, 206 are male and 62 are male and have a graduate degree.

xxMikexx [17]

Answer:

0.3009 is the probability that the applicant has graduate degree given he is a male.

Step-by-step explanation:

We are given he following in the question:

M: Applicant is male.

G: Applicant have a graduate degree

Total number of applicants = 450

Number of male applicants = 206

$n(M) = 206$

Number of applicants that are male and have a graduate degree = 62

$n(M\cap G) = 62$

$\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$P(M) = \dfrac{206}{450} = 0.4578$

$P(M\cap G) = \dfrac{n(M\cap G)}{n} = \dfrac{62}{450} = 0.1378$

We have to find the probability that the applicant has graduate degree given he is a male.

$P(G|M) = \dfrac{P(G\cap M)}{P(M)} = \dfrac{\frac{62}{450}}{\frac{206}{450}} = \dfrac{62}{206} = 0.3009$

Thus, 0.3009 is the probability that the applicant has graduate degree given he is a male.

5 0

3 years ago

Does anyone know how to solve this problem?

marissa [1.9K]

7? Is seems that the 4 is being added with +2 when it got bigger so I’m thinking the 5 would be the same thing +2 equals 7..that’s what I’m thinking

8 0

3 years ago

Suppose the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average

arsen [322]

Answer:

0.5372

Step-by-step explanation:

Given that the number of births that occur in a hospital can be assumed to have a Poisson distribution with parameter = the average birth rate of 1.8 births per hour.

Let X be the no of births in the hospital per hour

X is Poisson

with mean = 1.8

the probability of observing at least two births in a given hour at the hospital

= $P(X\geq 2)\\= 1-F(1)\\= 1-0.4628\\= 0.5372$

the probability of observing at least two births in a given hour at the hospital = 0.5372

4 0

3 years ago

What is the value of the expression x + (–y) when x = –2.3 and y = –5.9?

kramer

X + (-y) = -2.3 + ( - (-5.9) ) = -2.3 + 5.9 = 3.6

5 0

3 years ago

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