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adoni [48]
2 years ago
6

A group of people were asked if they had run a red light in the last year. 385 responded "yes", and 472 responded "no".

Mathematics
1 answer:
Dimas [21]2 years ago
3 0

Answer:

44.9%

Step-by-step explanation:

you can find the probability by dividing the total number of people by the number of people who have run a red light:

385/857

simplify into a decimal:

0.449

convert into a percent to get your answer:

44.9%

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Diane walked her dog 4/10 of a mile in 5 days this week. how far did diane walk her dog this week?
Charra [1.4K]

Answer:

.56 miles or 56/100 of a mile

Step-by-step explanation:

5 days = 4/10 = .4

7 days = ?

.4 / 5 * 7 = .56

.56 miles or 56/100 of a mile

7 0
3 years ago
С
serious [3.7K]

Answer:

10.6 oz

Step-by-step explanation:

X × 0.25 = 0.265

x= 2.65/0.25= 10.6 oz

8 0
2 years ago
Find the lateral surface area of the cylinder. Round your answer to the nearest tenth.
Svet_ta [14]

Answer:

A=2πrh

Step-by-step explanation:

A=2*22/7*4*9

=230

5 0
3 years ago
Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +
GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)  

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3)        ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2)        ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)    

On solving above equation for x    

(x+6) (x+3) =2(x+3) (2x+2)  

x+6 = 4x + 4    

x-4x = 4 – 6

x = \frac{2}{3}

Dimension of pool A

Length = x+6 = (\frac{2}{3}) +6 = 6.667m

Width = x +3 = (\frac{2}{3}) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2(\frac{2}{3}) + 6 = \frac{22}{3} = 7.333m

Width = 2x + 2 = 2(\frac{2}{3}) + 2 = \frac{10}{3} = 3.333m

Verifying the area:

Area of pool A = (\frac{20}{3}) x (\frac{11}{3}) = \frac{220}{9} = 24.44 meter

Area of pool B = (\frac{22}{3}) x (\frac{10}{3}) = \frac{220}{9} = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0
3 years ago
Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa
Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : \mu = 15.45

Standard deviation : \sigma = 13.70

The formula to calculate the z-score :-

z=\dfrac{x-\mu}{\sigma}

For x= 5 degrees

z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76

For x= 10 degrees

z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40

The P-value : P(-0.76

=0.3445783-0.2236273=0.120951\approx0.1210

In percent , 0.1210\times100=12.10\%

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

5 0
2 years ago
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