The graphs shown are of the form y = ax2. Which graph has the smallest value for a?

1 answer:

3 0

$y= x^{2}$ is a parabola (looks like the letter U).

The letter a represents the coefficient of $x^{2}$ and it controls two things (1) how wide or narrow the parabola is and (2) whether it is concave up (like a U) or concave down (like an up-side-down).

The absolute value of a (the number without the sign) controls how wide or narrow it is. If the absolute value is a fraction less than 1 the graph gets wider. The smaller the absolute value of the fraction the wider the graph gets.

If the absolute value of a is greater than 1 the graph gets narrower (it gets skinnier). The bigger the absolute value the narrower the graph.

So, if all the graphs look like a U (concave up) then the one with the smallest a is the one that is the widest.

The a also controls whether the graph is concave up or concave down. If a is negative

If a is negative the graph is concave down so any graph that is concave down has a smaller value of a than any graph that is concave up. However, if the graph is concave down the one with the smallest a would be the most narrow one.

So to find the one with the smallest a...
If they are all concave up (like a U) pick the widest one
and
If they are not all concave up pick the narrowest one that is concave down (looks like an upside down U)

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

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