This graph shows a proportional relationship.

KatRina [158]

Solve by Substitution
Answer:
x = -5
y = -2

5 0

1 year ago

At the start of his race, 86-kg Olympic 100m champion Usain Bolt pushes against the starting block, exerting an average force of

Leviafan [203]

1) Determine the horizontal force, Fx, exerted by the block on Usain

cos(20°) = Fx / F => Fx = F*cos(20) = 1700N*cos(20) = 1597.48 N

2) Determine the Impulse of that force on Usain.

I = F*t = 1597.48N * 0.32s = 511.19 N*s

3) Determine change in momentum of Usain, Δp

Δp = I = 511.19 N*s

4) Find change if velocity

Δp = Δ(mV) = mΔV => ΔV = Δp / m = 511.19 N*s / 86 kg = 5.94 m/s

Given the Usain started from rest, the velocity is ΔV - 0 = 5.94 m/s

Answer: 5.94 m/s

3 0

3 years ago

What is 275.9 × 10−4 in standard form?

lisov135 [29]

Answer:

-2755

Step-by-step explanation:

4 0

3 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

$z= \frac{X -\mu}{\sigma}$

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0

2 years ago

What is 90 divided by 12???? <br> pls help i need to get my grades up

Andrew [12]

Answer:

7.5

Step-by-step explanation:

cant u just g oogle it?

5 0

2 years ago

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