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2 years ago

7

Two points A and B whose heights differ by 10 metres are shown 1.2 cm apart horizontally on a map whose scale is 1: 10 000. a) W

ork out the actual horizontal distance between A and B.

2 answers:

PSYCHO15rus [73]2 years ago

6 0

Answer:

120 meters

Step-by-step explanation:

1.2 x 10 000 = 12000 cm = 120 meters

Snowcat [4.5K]2 years ago

3 0

120 meters is the answers

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Steven bought treats for his friends at the mall food court. Where m,n,p, and q are whole numbers, the amount Steven spent on th

Fed [463]

We have that the term from the expression that represents the amount of money Steven spent on cinnamon rolls is

X=3.19p

From the question we are told that

m,n,p, and q are whole numbers,

Expression Total amount spent =2.35m + 2.75n + 3.19p + 3.39q

Cost of cinnamon rolls=\$3.19

Generally the equation for the Money spent on cinnamon roll is mathematically given as

X=No of cinnamon roll * Cost of cinnamon roll

X=3.19*p

X=3.19p

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2 years ago

PLEASE HELP!!! I JUST NEED A STEP-BY-STEP!!!!!!

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Answer:

Step-by-step explanation:

$=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}x\sec^2(xy)dy\\=5\displaystyle\int_{0}^{\pi/6}dx\displaystyle\int_{0}^{2}\sec^2(xy)d(xy)\\=5\displaystyle\int_{0}^{\pi/6}dx\tan(xy)|_{y=0}^{y=2}$

$=5\displaystyle\int_{0}^{\pi/6}\tan(2x)dx\\=-\frac{5}{2}\ln\cos(2x)|_{0}^{\pi/6}\\=-\frac{5}{2}[\ln\cos(\pi/3) - \ln\cos(0)]\\$

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1 year ago

A bread recipe calls for 1 teaspoon of yeast for every 2

VashaNatasha [74]

C - T = T would be the correct answer

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3 years ago

Can someone help please!! Provide workout please!!!

mamaluj [8]

Answer:

it is 55%

Step-by-step explanation:

I think it is 55 %

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2 years ago

Read 2 more answers

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

2 years ago