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3 years ago

(4.76) × (12.5) simplified

Mathematics

2 answers:

densk [106]3 years ago

6 0

Answer:

I believe the answer is 59.5.

Step-by-step explanation:

If I understood this problem correctly, you'd have to multiply 4.76 times 12.5. As an answer, I got 59.5.

dsp733 years ago

4 0

Answer:

59.5

Step-by-step explanation:

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A study investigated the effectiveness of meditation training in reducing trait anxiety. The study evaluated subjects trait anxi

Setler79 [48]

Answer:

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 27 - 1 = 26

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.7787.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.7787\frac{6.9}{\sqrt{27}} = 3.7$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.

The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

6 0

2 years ago

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 2

Shalnov [3]

Answer:

The expected winnings for a person buying 1 ticket is -0.2.

Step-by-step explanation:

Given : A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each. If 20000 tickets are sold at 25 cents each, find the expected winnings for a person buying 1 ticket.

To find : What are the expected winnings?

Solution :

There are one first prize, 2 second prize and 20 third prizes.

Probability of getting first prize is $\frac{1}{20000}$

Probability of getting second prize is $\frac{2}{20000}$

Probability of getting third prize is $\frac{20}{20000}$

A raffle offers a first prize of $1000, 2 second prizes of $300, and 20 third prizes of $10 each.

So, The value of prizes is

$\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10$

If 20000 tickets are sold at 25 cents each i.e. $0.25.

Remaining tickets = 20000-1-2-20=19977

Probability of getting remaining tickets is $\frac{19977}{20000}$

The expected value is

$E=\frac{1}{20000}\times 1000+\frac{2}{20000}\times 300+\frac{20}{20000}\times 10-\frac{19977}{20000}\times 0.25$

$E=\frac{1000+600+200-4994.25}{20000}$

$E=\frac{-3194.25}{20000}$

$E=-0.159$

Therefore, The expected winnings for a person buying 1 ticket is -0.2.

3 0

2 years ago

A rectangle has a length of 2 m and a width of 40 cm what is the perimeter of the rectangle ( show your work)

sweet [91]

Answer:

0.05 im pretty sure to my calcutor

Step-by-step explanation:

4 0

3 years ago

6(6x−3)= helpppppppppppppp distibutive property again

bagirrra123 [75]

Answer:

6(6x-3)

=36x-18

Step-by-step explanation:

6(6x-3(

multiply 6 x 6 and 6 x 3

36x - 18

5 0

3 years ago

Find the midpoint between −18 and 52.

Mama L [17]

Finding the midpoint between two points on a line involves three steps:

1. Find the distance between the two points
2. Find half of that distance
3. Either add that half to the lowest point, or subtract it from the highest point.

To find the distance between 52 and -18, we can take the absolute value of their difference. The reason we take the absolute value is to force the sign to be positive, since distances can never be negative. -18 - 52 = -70, and |-70| = 70.

Second, we find half of that distance, which in this case would be 70/2 = 35.

Finally, we add that value to -18 to find a midpoint of -18 + 35 = 17.

5 0

3 years ago