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2 years ago

What is the probability of rolling an odd number or rolling a 2 on a six sided die? *

Mathematics

1 answer:

OLga [1]2 years ago

4 0

Answer: 2/3

Step-by-step explanation:

First, find the probability of rolling an odd number. The odd numbers on a 6-sided die are 1, 3, and 5. This is a probability of 3/6.

Now, find the probability of rolling a 2. There's only one 2 on a die, so the probability is 1/6.

Next, add these together. 3/6 + 1/6 = 4/6. This can be reduced by dividing both the numerator and denominator by 2. This results in a final probability of 2/3.

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The Waist to Hip Ratio (WHR) is a ratio commonly expressed as a decimal that has been shown to be a good predictor of possible c

Jet001 [13]

40?
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3 years ago

Rectangle EFGH is the image of rectangle ABCD after a sequence of transformations. Which statement describes the transformation

nasty-shy [4]

Answer:

C. Rectangle ABCD was dilated with center (0, 0) and a scale factor of 1/2 followed by a translation right 3.5 units and down 8 units.

Step-by-step explanation:

See figure for intermediate steps.

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3 years ago

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miss watson runs a distance of 200m in 25 seconds work out the average speed of miss watson in metres per second

Kobotan [32]

Answer:

8 meters per second

Step-by-step explanation:

4 0

2 years ago

Lincoln went to the grocery store and purchased cans of soup and frozen dinners. Each can of soup has 250 mg of sodium and each

Ludmilka [50]

We can solve with a system of equations, and use c for the amount of cans of soup and f for the amount of frozen dinners.

The first equation will represent the amount of sodium. We know the (sodium in one can times the number of cans) plus (sodium in one frozen dinner times the number of dinners) is the expression for the total sodium. We also know the total sodium is 4450, so:

250c + 550f = 4450

The second equation is to find how many of each item are purchased:

c + f = 13

Solve for c in the second equation:

c = 13 - f

Plug this in for c in the first equation:

250(13-f) + 550f = 4450

3250 - 250f + 550f = 4450

300f = 1200

f = 4

Now plug the value for f into the second equation:

c + 4 = 13

c = 9

The answer is 9 cans of soups and 4 frozen dinners.

6 0

3 years ago

Read 2 more answers

Special right triangles.

Oliga [24]

QUESTION 9

The given right angle triangle has its two legs equal to $x\:cm$ and the hypotenuse is $2\:units$ .

Using the Pythagoras theorem, we have

$x^2+x^2=2^2$

This implies that;

$2x^2=2^2$

$2x^2=4$

$x^2=2$

$x=\sqrt{2}$

$x=1.414$

$x\approx1.4\:units$

QUESTION 10

The given right angle triangle has its two legs equal to $x\:units$ and the hypotenuse is $7\:units$ .

Using the Pythagoras theorem, we have

$x^2+x^2=7^2$

This implies that;

$2x^2=7^2$

$2x^2=49$

$x^2=24.5$

$x=\sqrt{24.5}$

$x=4.95\:units$

QUESTION 11

Sam divided the square backyard into two sections along the 40ft diagonal.

Let one of the sides of the garden be $x\:units$ , then the other side of the garden is also $x\:units$ since it was a square backyard.

The diagonal is the hypotenuse which is 40ft and the two legs are $x\:units$ each.

Applying the Pythagoras Theorem, we have;

$x^2+x^2=40^2$

$2x^2=1600$

$x^2=800$

$\Rightarrow x=\sqrt{800}$

$\Rightarrow x=28.28$

The length of one side of the garden is approximately 28cm.

QUESTION 12

The hypotenuse of Nicole's right triangular support is 92cm long.

It was given that the two lengths of the triangular support are of equal length.

Let the two lengths of the triangular support be $l\:cm$ each.

We then apply the Pythagoras theorem to obtain;

$l^2+l^2=92^2$

$\Rightarrow 2l^2=8464$

$\Rightarrow l^2=4232$

$\Rightarrow l=\sqrt{4232}$

$\Rightarrow l=65.05cm$

Therefore Nicole needs $l+l=65.05cm+65.05cm\approx130cm$ of wood to complete the support.

The correct answer is A.

QUESTION 13

A 45-45-90 triangle is an isosceles right triangle.

Let the length of the two legs be $m\:inches$ each.

It was given that the hypotenuse of this triangle is $12\:inches$ .

Applying the Pythagoras Theorem, we have;

$m^2+m^2=12^2$

$\Rightarrow 2m^2=144$

$\Rightarrow m^2=72$

$\Rightarrow m=\sqrt{72}$

$\Rightarrow m=8.49$

The length of one leg of the hypotenuse is approximately 8.5 inches to 1 decimal place.

QUESTIONS 14.

It was given that the distance from home plate to first base is 90 feet.

Since the baseball field form a square, the length of the baselines are all 90ft.

We want to calculate the distance from home plate to 2nd base,which is the diagonal of the square baseball field.

Let this distance be $d\:ft$ , then using the Pythagoras theorem;

$d^2=90^2+90^2$

$d^2=8100+8100$

$d^2=16200$

$\:d=\sqrt{16200}$

$\:d=127.279$

Therefore the distance from home plate to second base is 127 ft to the nearest foot.

QUESTION 15.

The distance from third base to first base is also 127 ft to the nearest foot.

The reason is that the diagonals of a square are equal in length.

3 0

3 years ago