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2 years ago

If you roll a number cube 12 times, about how many times would you expect to roll a 5 or 6

Mathematics

1 answer:

Marrrta [24]2 years ago

6 0

You have a 1/3 chance each time you role So 1/3 x 12 = 4 times that you would role a 5 or a 6

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What is the width of a rectangle with an area of 5/8 inches2 and a length of 1o inches

Nata [24]

I hope this helps you

Area=length.width

Area=1.5/8

Area=5/8

5 0

3 years ago

Use the figure to the right to find the value of PT. T is the midpoint of PQ.

Monica [59]

Answer:

$PQ = 42$

Step-by-step explanation:

Given

$PT = 3x + 3$

$TQ = 5x - 9$

Midpoint, T

Required

Determine PT

First, the value of x has to be calculated;

Since T is the midpoint, then:

$PT = TQ$

This implies

$3x + 3 = 5x - 9$

Solve for x

$3x - 5x = -3 - 9$

$-2x = -12$

$x = -12/-2$

$x = 6$

Length PQ can then be calculated using:

$PQ = PT + TQ$

$PQ = 3x +3 + 5x -9$

Collect Like Terms

$PQ = 3x+ 5x +3 -9$

$PQ = 8x -6$

Substitute 6 for x

$PQ = 8 * 6 - 6$

$PQ = 48 - 6$

$PQ = 42$

6 0

3 years ago

Which inequality represents the situation? My dogs weight is greater than 80 pounds.

Nata [24]

Answer:Dog's weight> 80 pounds

Step-by-step explanation:Dog's weight is greater than 80 pounds

8 0

3 years ago

What is the exact value of x?

Nadya [2.5K]

Exact value of x will be 8 x sin(60)

Exact value of sin 60 = √3/2

So √3/2 x 8= 8 √3/2

Can be simplified to 4 √3

8 0

2 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

2 years ago