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pychu [463]
3 years ago
13

The graph below shows the height of a projectile t seconds after it is launched. If acceleration due to gravity is –16 ft/s2, wh

ich equation models the height of the projectile correctly?

Mathematics
1 answer:
nika2105 [10]3 years ago
5 0

Answer:

  see attached

Step-by-step explanation:

The x-coordinate of the vertex is given by ...

  x = -b/(2a)

for f(x) = ax^2 +bx +c.

We see that the vertex is at x = 1, and we are given a = -16. Then we can find "b" from ...

  1 = -b/(2(-16))

  32 = b

We know the y-intercept of the equation is 5 (the starting height), so the equation must be ...

  h(t) = -16t^2 +32t +5 . . . . . matches the last choice

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What decimal is smaller .52 or .25
Liono4ka [1.6K]

Answer:

0.25

Step-by-step explanation:

0.52 is like 2 quarters

0.25 is 1 quarter

So, 1 < 2

Have a nice day! :)

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3 years ago
Simplify the expression -|-5 x (-7)|
Andrei [34K]

Answer:

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7 0
3 years ago
What is the answer I'm very confused???????
Tanzania [10]
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6 0
3 years ago
Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

8 0
3 years ago
3. List all the coefficients and constants
7nadin3 [17]
The coefficient is 4
The constants are 2 and 3
5 0
3 years ago
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