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3 years ago

Plssss help I’ll mark u brainliest!!!

Mathematics

1 answer:

Nutka1998 [239]3 years ago

4 0

Answer:

The answer to the question is C

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A person gets on and off a bathroom scale four times. The four readings (in pounds) are 148, 151, 150, and 152. Each time after

Ganezh [65]

Answer:

See explanation below

Step-by-step explanation:

Since each time after the person gets off the scale, the reading is 2 lb the person's weight must be near the mean of

148-2, 151-2, 150-2, 152-2; that is to say, near the mean of 146, 149, 148, 150 = (146+149+148+150)/4 = 148.25

We could estimate the uncertainty as the standard error SE

$\bf SE=\frac{s}{\sqrt{n}}$

where

s = standard deviation of the sample

n = 4 sample size.

Computing s:

$\bf s=\sqrt{\frac{(146-148.25)^2+(149-148.25)^2+(148-148.25)^2+(150-148.25)^2}{4}}=1.479$

So, the uncertainty is 1.479/2 = 0.736

It is not possible to estimate the bias, since it is the difference between the true weight and the mean, but we do not know the true weight.

5 0

2 years ago

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

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2 years ago

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Hunter-Best [27]

Answer:

7,515

Step-by-step explanation:

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3 years ago

Please help with 47-50

Tems11 [23]

Answer:

-3

Step-by-step explanation:

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3 years ago

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How you this and the answers

gregori [183]

Answer:

Step-by-step explanation:

? whats the question

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2 years ago