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3 years ago

7

The world record for the greatest temperature range recorded in one day occurred in Browning, Montana, in 1916. The temperature

fell from 44°F to –56°F. What was the temperature change that day?

1 answer:

balu736 [363]3 years ago

3 0

Answer:

The temperature dropped 100 degrees.

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What is the y-coordinate of the solution to this system of equations?<br> 5x-8y=1<br> 3x+6y=-21

Vanyuwa [196]

5x-8y=1...(1)
3x+6y=-21...(2)

(1)*3:
15x-24y=3...(3)

(2)*5:
15x+30y=-105...(2)

(3)-(2):
15x-24y-(15x+30y)=3-(-105)
15x-24y-15x-30y=108
-54y=108
y=-2

6 0

3 years ago

The data given below contain the state cigarette taxâ€‹ (in $) for all 20 regions of a country.

g100num [7]

Answer:

(a) $\bar x= 1.583$ -- Population Mean

(b) $s = 1.032$ --- Population standard deviation

(c) See Explanation

Step-by-step explanation:

Given:

Cigarette tax for 20 regions

Solving (a): The population mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

$\sum x = 1.36 + 1.70 + 2.50 + 0.45 + 1.18 + 0.64 + 3.46 + 0.57 + 2.00 + 0.80 + 1.60 +0.98 + 0.36 + 2.24 + 4.35 + 0.62 + 2.70 + 1.78 + 1.53 + 0.84$

$\sum x = 31.66$

$n = 20$

So, we have:

$\bar x= \frac{31.66}{20}$

$\bar x= 1.583$

Solving (b): The population standard deviation

This is calculated as:

$s = \sqrt{\frac{\sum( x - \bar x)^2}{n}$

$\sum (x -\bar x)^2 = (1.36 - 1.583)^2 + (1.70 - 1.583)^2+ (2.50 - 1.583)^2+ (0.45 - 1.583)^2+ (1.18 - 1.583)^2+ (0.64 - 1.583)^2+ (3.46 - 1.583)^2+ (0.57 - 1.583)^2+ (2.00 - 1.583)^2+ (0.80 - 1.583)^2+ (1.60 - 1.583)^2+(0.98 - 1.583)^2+ (0.36 - 1.583)^2+ (2.24 - 1.583)^2+ (4.35 - 1.583)^2+ (0.62 - 1.583)^2+ (2.70 - 1.583)^2+ (1.78 - 1.583)^2+ (1.53 - 1.583)^2+ (0.84- 1.583)^2$

$\sum (x -\bar x)^2 = 21.29222$

So:

$s = \sqrt{\frac{21.2922}{20}$

$s = \sqrt{1.06461}$

$s = 1.032$

Solving (c):

Population mean tells the average amount while the standard deviation represents the spread from the calculated mean

Option (4) is correct

6 0

3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

2x^2-13x+20=2x <br> 2<br> −13x+20

marta [7]

Answer:

Factor by grouping

( 2 x + 5 ) ( x + 4 )

6 0

3 years ago

Simplify the expression. 2(x-3)+4y-2(x-y-3)+5

Rasek [7]

2(x - 3) + 4y - 2(x - y - 3) + 5 =
2x - 6 + 4y - 2x + 2y + 6 + 5 =
6y + 5 <== ur x terms cancel and ur 6's cancel

3 0

3 years ago

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