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3 years ago

I want yo nkow how to simplify

1 answer:

5 0

Remove parentheses by multiplying factors.
use exponent rules to remove parentheses in terms with exponents.
combine like terms by adding coefficients.
combine the constants.

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A football is kicked toward an end zone with an initial vertical velocity of 30 ft/s. The function h(t) = -16+ 30t models the he

Ulleksa [173]

Answer:

A. The football does not reach a height of 15ft

Step-by-step explanation:

Given

$h(t) = -16t^2 + 30t$

Required

Determine which of the options is true

The option illustrates the height reached by the ball.

To solve this, we make use of maximum of a function

For a function f(x)

Such that:

$f(x) = ax^2 + bx + c$ :

$f(\frac{-b}{2a}) = maximum/minimum$

i.e we first solve for $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for x in $f(x) = ax^2 + bx + c$

In our case:

First we need to solve $\frac{-b}{2a}$

Then substitute $\frac{-b}{2a}$ for t in $h(t) = -16t^2 + 30t$

By comparison:

$b = 30$

$c = -16$

$\frac{-b}{2a} = \frac{-30}{2 * -16}$

$\frac{-b}{2a} = \frac{-30}{-32}$

$\frac{-b}{2a} = \frac{30}{32}$

$\frac{-b}{2a} = \frac{15}{16}$

Substitute $\frac{15}{16}$ for t in $h(t) = -16t^2 + 30t$

$h(\frac{15}{16}) = -16(\frac{15}{16})^2 + 30(\frac{15}{16})$

$h(\frac{15}{16}) = -16(\frac{225}{256}) + \frac{450}{16}$

$h(\frac{15}{16}) = -(\frac{225}{16}) + \frac{450}{16}$

$h(\frac{15}{16}) = \frac{-225 + 450}{16}$

$h(\frac{15}{16}) = \frac{225}{16}$

$h(\frac{15}{16}) = 14.0625$

This implies that the maximum height reached is 14.0625ft.

So, the option that answers the question is A because $14.0625 < 15$

8 0

2 years ago

Consider the quadratic function f(x) = x2 – 5x + 6.

Deffense [45]

The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation. Therefore,

f(x) = x2 – 5x + 6
a = 1
b = -5
c = 6

8 0

3 years ago

Rewa Delta Union Rugby CEO has become concerned about the slow pace of the rugby games played in the current union rugby, fearin

ozzi

Answer:

We conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

Step-by-step explanation:

We are given that Before the meeting, the mean duration of the 15-sided rugby game time was 3 hours, 5 minutes, that is, 185 minutes.

A random sample of 36 of the 15-sided rugby games after the meeting showed a mean of 179 minutes with a standard deviation of 12 minutes.

Let $\mu$ = mean duration of 15-sided union rugby games after the meeting.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 185 minutes {means that the mean duration of 15-sided union rugby games has increased or remained same after the meeting}

Alternate Hypothesis, $H_A$ : $\mu$ < 185 minutes {means that the mean duration of 15-sided union rugby games has decreased after the meeting}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean duration of 15-sided union rugby games = 179 min

s = sample standard deviation = 12 minutes

n = sample of 15-sided rugby games = 36

So, the test statistics = $\frac{179-185}{\frac{12}{\sqrt{36} } }$ ~ $t_3_5$

= -3

The value of t test statistics is -3.

Now, at 1% significance level the t table gives critical value of -2.437 at 35 degree of freedom for left-tailed test.

Since our test statistic is less than the critical value of t as -3 < -2.437, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean duration of 15-sided union rugby games has decreased after the meeting.

6 0

3 years ago

Can someone solve this please 9g = 36

Goryan [66]

You divide 9 on both sides than it would be g=4

3 0

3 years ago

Read 2 more answers

A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random sample of 5 items is chosen to be inspected.

dimulka [17.4K]

Answer:

a. 39.55%

b. 44.02%

Step-by-step explanation:

We have the following data:

n = 5

x = 1

p = 5/20 = 0.25

to. If the sampling is done with replacement.

We apply the binomial distribution formula, which is as follows:

P = nCx * (p ^ x) * ((1-p) ^ (n-x))

Where nCx, is a combination, and is equal to:

nCx = n! / x! * (n-x)!

replacing we have:

5C1 = 5! / 1! * 4! = 5

replacing in the main formula:

P = 5 * (0.25 ^ 1) * ((1- 0.25) ^ (5-1))

P = 0.3955

that is, without replacing the probability is 39.55%

b. if the sampling is done without replacement.

Here it is a little different from the previous one, but what you should do is calculate three cases,

the first was the one at point a, when n = 5 and x = 1

5C1 = 5! / 1! * 4! = 5

the second is when n = 20 and x = 5, this is all possible scenarios.

20C5 = 20! / 5! * 15! = 15504

and the third is when n = 15 (20-5) and x = 4 (5-1), which corresponds to the cases when none were damaged

15C4 = 15! / 4! * 11! = 1365

In the end, it would be:

P = (5C1 * 15C4) / 20C5

Replacing:

P = 5 * 1365/15504

P = 0.4402

Which means that without replacing the probability is 44.02%

7 0

3 years ago