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3 years ago

I need help with this question.

Mathematics

1 answer:

Ivanshal [37]3 years ago

8 0

Answer:

Step-by-step explanation:

f(x-2) means that x is happening sooner or a shift to the left and

+4 means that the whole function moves up 4.

The 1st choice looks good

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FOR 15 POINTS I WILL GIVE BRAINLEST IF YOU ARE CORRECT! What is the solution to the system that is created by the equation y = 2

goblinko [34]

The solution point is (4,2).

Step-by-step explanation:

Equation of straight line passing through the points (0,0) and (4,2) is given by

⇒ 2y = x ............ (1)

And the other straight line is y = - x + 6 .............. (2)

Now, solving equations (1) and (2) we get, y = - 2y + 6

⇒ 3y = 6

⇒ y = 2

And from equation (1) we get, x = 2y = 4

Therefore, the solution point is (4,2). (Answer)

6 0

3 years ago

Read 2 more answers

g In a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained

Marat540 [252]

Answer:

95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].

Step-by-step explanation:

We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.

Firstly, the Pivotal quantity for 95% 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 residents 65-years or older who had obtained a flu shot = $\frac{28}{111}$ = 0.25

n = sample of residents 65-years or older = 111

p = population proportion of residents who were getting the flu shot

Here for constructing 95% confidence interval we have used One-sample z test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.25-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ , $0.25+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{111} } }$ ]