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4 years ago

What is the probability of randomly selecting an odd number from the numbers 1 -15?

Mathematics

1 answer:

Scilla [17]4 years ago

7 0

Answer:

7/15

Step-by-step explanation:

So basically, you need to count the numbers (3,5,7,9,11,13,15) those are 7 numbers in total, so that's 7/15, if you want a percentage its 47% (rounded)

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Can someone explain how to find the answer to x

OLga [1]

Answer:

130

Step-by-step explanation:

X is the alternate corresponding angle to 50. 50+y=180

y=130

x and y = the same thing so 130 is the value of x

4 0

3 years ago

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The local toy store has a bin of toy vehicles for sale. The bin holds b bikes and c cars. If the store sells 1/3 of the vehicles

LekaFEV [45]

1/3b+1/3c is the expression assuming bikes are counted as vehicles

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3 years ago

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I KNOW THE ANSWER IS NOT 7. DO NOT RESPOND WITH 7.

schepotkina [342]

Answer:

what do you mean it's not 7 it is 7

Step-by-step explanation:

The value of x is 7

The given parameters are:

\angle BAE = 9x + 2∠BAE=9x+2

\angle BAD = 130^o∠BAD=130

Line AE is the angle bisector of \angle BAD∠BAD

This means that:

\angle BAD = 2 \times \angle BAE∠BAD=2×∠BAE

So, we have:

130 = 2 \times (9x + 2)130=2×(9x+2)

Divide both sides by 2

65 = 9x + 265=9x+2

Subtract 2 from both sides

63 = 9x63=9x

Divide both sides by 9

7 = x7=x

Rewrite as:

x = 7x=7

Hence, the value of x is 7

3 0

2 years ago

Entomologists have discovered that a linear relationship exists between the rate of chirping of crickets of a certain species an

vekshin1

Answer:

a) $y=4x-160$

b) $y=4(102)-160=248\frac{chirps}{min}$

Step-by-step explanation:

We know that theres a linear relationship between the rate of the chirping of crickets and the air temperature.

The equation of a line is:

$y=mx+b$

So, let's name our variables

x= Temperature

y=Rate of the chirping

First of all, we need to find the slope with the two given points

$x_{1}=$ 60ºF , $y_{1} =80\frac{chirps}{min}$

$x_{2} =$ 80ºF, $y_{2} =160\frac{chirps}{min}$

By,

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} } =\frac{160-80}{80-60}$

$m=4$

Now, the equation between the air temperature and the number of chirps is:

$y-y_{1} =m(x-x_{1} )$

$y-80=4(x-60)$

Solving for y,

a) $y=4x-160$

b) To calculate the rate at which the crickets chirp when the temperature is 102 ºF we need to evaluate y(102)

$y=4(102)-160=248\frac{chirps}{min}$

5 0

3 years ago

BRAINLIEST!!

Kaylis [27]

Converse of alt. interior angles theorum

7 0

3 years ago

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