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Irina-Kira [14]

3 years ago

12

Distance on a street map is proportional to actual distance when the distance on the map is (1)/(2) cm, the actual distance is 3

km. shannas house is 16(1)/(2)km from the riding stables what is the distance on the map
a) 2 (3)/(4) cm
b) 5 (1)/(2) cm
C) 6cm
D) 11 cm

1 answer:

bixtya [17]3 years ago

6 0

Answer:

b) 5 (1)/(2) cm

Step-by-step explanation:

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Which of the following ratios is equivalent to 5 to 20

morpeh [17]

<h2><em>which of the following ratios is equivalent to 5 to 20</em></h2>

<em>10:40</em>

Step-by-step explanation:

<em>because 5/20=1/4 and 10/40 is also equal to 1/4.</em>

<em>hope </em><em>it</em><em> helps</em>

4 0

3 years ago

A football team has a probability of 0.76 of winning when playing any of the other four teams in its conference. If the games ar

Deffense [45]

Answer:

0.33362176

Step-by-step explanation:

Probability is the ratio of the number of possible outcomes to the number of total outcomes.

Given that a football team has a probability of 0.76 of winning when playing any of the other four teams in its conference, the probability that the team wins all its conference games

= 0.76 * 0.76 * 0.76 * 0.76

= 0.33362176

This is the probability that the teams wins the 4 matches played against the other teams in the conference.

3 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2700 grams and a standard deviation of 900

rosijanka [135]

Answer:

The 33 week gestation period baby has a zscore of -0.47.

The 41 week gestation period baby has a zscore of -0.94.

The 41 week gestation period baby weighs less relative to the gestation period.

Step-by-step explanation:

Normal model problems can be solved by the zscore formula.

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

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

The zscore represents how many standard deviations the value of X is above or below the mean $\mu$ . This means that the baby with the lowest Zscore is the one who weighs relatively less to the gestation period.

33 week gestation period baby:

Babies born after a gestation period of 32 to 35 weeks have a mean weight of 2700 grams and a standard deviation of 900 grams, so $\mu = 2700, \sigma = 900$ .

A 33-week gestation period baby weighs 2275 grams. So $X = 2275$ .

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

$Z = \frac{2275 - 2700}{900}$

$Z = -0.47$

41 week gestation period baby:

Babies born after a gestation period of 40 weeks have a mean weight of 3200 grams and a standard deviation of 450 grams, so $\mu = 3200, \sigma = 450$

A 41-week gestation period baby weighs 2775 grams, so $X = 2775$ .

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

$Z = \frac{2775 - 3200}{450}$

$Z = -0.94$

The 41 week gestation period baby weighs less relative to the gestation period, since he has a lower zscore.

8 0

3 years ago

Find the perimeter and area of the rectangle.

Inessa [10]

Answer:

P = 66 ft

A = 216 sq ft

Step-by-step explanation:

Perimeter:

length + length + width + width

24 + 24 + 9 + 9 = 66

Area:

24 x 9 = 216

4 0

3 years ago

Read 2 more answers