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3 years ago

Which statement is true for the scatter plot? The data show a positive correlation.

Mathematics

1 answer:

il63 [147K]3 years ago

3 0

Answer:

The data shows a postive correlation.

Step-by-step explanation:

You might be interested in

Clara pays $76 for 8 guitar lessons. How much would 10 guitar lessons cost? Enter your answer in the box.

WARRIOR [948]

You would divide 76 by 8. That would give you 9.5. Then you multiply by 10 and that gives you 95. Hope this helps (;

5 0

4 years ago

Suppose that only 65% of all drivers in a certain state wear a seat belt. a random sample of 80 drivers is selected. what is the

jonny [76]

Answer:

Pr(X >42) = Pr( Z > -2.344)

= Pr( Z< 2.344) = 0.9905

Step-by-step explanation:

The scenario presented can be modeled by a binomial model;

The probability of success is, p = 0.65

There are n = 80 independent trials

Let X denote the number of drivers that wear a seat belt, then we are to find the probability that X is greater than 42;

Pr(X > 42)

In this case we can use the normal approximation to the binomial model;

mu = n*p = 80(0.65) = 52

sigma^2 = n*p*(1-p) = 18.2

Pr(X >42) = Pr( Z > -2.344)

= Pr( Z< 2.344) = 0.9905

7 0

3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

5 0

3 years ago

283838383838382828 + 83838383888282ab

natta225 [31]

2.8392222e+17

Step-by-step explanation:

283838383838382828 + 83838383888282=2.8392222e+17

5 0

3 years ago

A book sold 31,500 copies in its first month of release. Suppose this represents 8.9% of the number of copies sold to date. How

kogti [31]

Answer:

35,393,258

Step-by-step explanation

$\frac{31,500}{x} =\frac{0.089}{100}$

$0.089x=3,150,000$

$x= 35,393,258.427$

Rounded to the nearest whole number, that is 35,393,258.

7 0

3 years ago