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Nina [5.8K]
3 years ago
13

You are testing fertilizers on plants. One plant is 11 cm tall and growing at a rate of 2 cm/day. Write an equation that can be

solved to find the number of days, d, it will take for the plant to be 65 cm tall.
Mathematics
1 answer:
Vlada [557]3 years ago
4 0
I would consider for you to use Socratic
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The equation y=x is a function. True or False?
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false

Step-by-step explanation:

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Algebra solve each equation (30 points)
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A flagpole 1400 cm tall is how many meters tall?
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Of the 10,000 participants in the Cooper Run/Walk, 42% were runners and one-third of the runners were female. Sally, a female ru
Aleks [24]

Answer: 140 female runners

Step-by-step explanation:

First find out the number of runners:

= 42% * 10,000

= 4,200 runners

Then find out the number of female runners:

= 4,200 * 1/3

= 1,400 female runners

Sally finished faster than 90% of the female runners which means that 10% finished faster or at the same time as her.

= 10% * 1,400

= 140 female runners

5 0
3 years ago
The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
Rainbow [258]

Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

8 0
3 years ago
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