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2 years ago

Which of the following linear systems has a solution (5,0) SOMEONE PLS HELPPP

Mathematics

2 answers:

crimeas [40]2 years ago

4 0

D is the answer. Have a nice day

Korvikt [17]2 years ago

3 0

Answer:

Step-by-step explanation:

out of the choices

A) 2x-y=-2

3x+2y=-10

B) x+4y=3

2x+5y=3

C) 2x+3y=5

x-4y=-14

D) 3x+4y=15

x+y=5

only the system in D has a solution of x=5 and y=0

solve by substitution x+y=5 so x=5-y

plug into the other equation so 3(5-y)+4y=15

distribute 15-3y+4y=15 then minus 15 on both sides and combine like terms

so y=0 then put that value back into the other equation

so x+y=5 equals x+0=5 so x=5 this means the solution to the system of equations is the point (5,0)

Hope this helps! :)

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One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]