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2 years ago

11

Suppose a soccer goalie punted the ball in such a way as to kick the ball as far as possible down the field. The height of the b

all above the field can be approximated by the function below where y represents the height of the ball (in yards) and x represents the horizontal distance (in yards) down the field from where the goalie kicked the ball. $y=0.017x^{2} +0.98x+0.33$
How far away did the ball land? Estimate to 2 places after the decimal.

1 answer:

Alexxx [7]2 years ago

8 0

Answer:

57.98 yards

Step-by-step explanation:

When the ball is on the ground, the height is $y=0$ , therefore:

$y=-0.017x^2+0.98x+0.33$

$0=-0.017x^2+0.98x+0.33$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-0.98\pm\sqrt{0.98^2-4(-0.017)(0.33)}}{2(-0.017)}$

$x_1=57.98184919$

$x_2=-0.3347903693$

Since distance cannot be negative, the ball landed 57.98 yards down the field from where the goalie kicked the ball.

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Answer:

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b

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Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

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So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

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$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

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$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

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Let us assume that $F = \= X - \= Y$

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Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

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