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4 years ago

15

St. Louis, Missouri and Washington, D.C. Have very similar latitudes. (Latitude measures how far north or south a place is, whic

h in turn affects the temperature.) The box plots below show the daily high temperatures (in degrees Celsius) that were recorded during July 201320132013 in each city. Which pieces of information can be gathered from these box plots?

2 answers:

igor_vitrenko [27]4 years ago

5 0

Answer:

The Answer is multiple choice and the answer is B. Only B.

Step-by-step explanation:

In-s [12.5K]4 years ago

4 0

Answer: the answer is B

Step-by-step explanation: this is because Washington has a higher median.

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I need help i dont know the answers

Arlecino [84]

Answer: Change the number: 36

To:48

Step-by-step explanation:

12x2 is 24 but 24 x 2 isn’t 36

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3 years ago

Janet has 3 sweatshirts, 5 t-shirts, and 2 pairs of pants that can be worn together. If each outfit has a sweatshirt, t-shirt, a

RideAnS [48]

Answer:

30 outfits

Step-by-step explanation:

First, come up with all the possibilties for one of the sweatshirts. So...

S11

S12

S21

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S31

S32

S41

S42

S51

S52

Count up all of those and then multiply it by three

10 * 3 = 30

Have a wonderful day!

4 0

3 years ago

The Stinker station raised the price on a gallon of gas from $2.45 to $2.63. What is the percent of increase on the price hike?

Mars2501 [29]

Answer:$5.00

Step-by-step explanation: ADDD

6 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

BRAINLEST IF CORRECT!!!!

Aleksandr-060686 [28]

x + 133 = 180

x = 180 - 133 (Linear Pair)

x = 47

Therefore x° = 47°

Answered by Gauthmath must click thanks and mark brainliest

7 0

3 years ago

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