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3 years ago

what is the rate of change and initial value for this graph?

Mathematics

1 answer:

MAXImum [283]3 years ago

8 0

Answer:

4/7

Step-by-step explanation:

the rate of change (aka slope) is the change in y over change in x and as y goes up 4 x goes up 7 so the rate of change is 4/7

have a great day :D

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Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago

Solve 732, 178 167-542, 137=

12345 [234]

732, 178 167-542, 137
= 731, 636, 030

4 0

4 years ago

HELP.......................

ira [324]

1. 1 2. 1 1/43.1 1/4

7 0

3 years ago

Y = x + 3 2y = 2x + 6

Marina CMI [18]

Answer:

x=0

y=3

Step-by-step explanation:

2(x+3)=2x+6, substituting the first equation in the second

2x+6=2x+6, rearranging the equations

2x-2x=6-6

0=0

y=0+3

y=3

6 0

4 years ago

Let us suppose that one of the fireworks is launched from the top of the building with an initial upward velocity of 150 ft/s an

ivanzaharov [21]

Answer:

9.6 seconds

Step-by-step explanation:

The equation for projectile motion in feet is $h(t)=-16t^2+v_0t+h_0$ where $v_0$ is the initial velocity in ft/s and $h_0$ is the initial height in feet.

We are given that the initial upward velocity is $v_0=150ft/sec$ and the initial height is $h_0=35ft$ . Thus, plugging these values into our equation, we get $h(t)=-16t^2+150t+35$ as our equation for the situation.

The firework will land, assuming it doesn't explode, when $h(t)=0$ , thus:

$h(t)=-16t^2+150t+35\\\\0=-16t^2+150t+35\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-150\pm\sqrt{150^2-4(-16)(35)}}{2(-16)}\\ \\t=\frac{-150\pm\sqrt{22500+2240}}{-32}\\\\t=\frac{-150\pm\sqrt{24740}}{-32}\\\\t=\frac{-150\pm2\sqrt{6185}}{-32}\\ \\t=\frac{75\pm\sqrt{6185}}{16}\\\\t_1\approx-0.2\\\\t_2\approx9.6$

Since time can't be negative, then the firework will land after 9.6 seconds, assuming it doesn't explode at the time of landing.

3 0

3 years ago

￼what is the rate of change and initial value for this graph?

what is the rate of change and initial value for this graph?