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3 years ago

446,221 to the nearest ten

2 answers:

8 0

To find the nearest tenth, first you have to know where it is, the tenths place is always the last number , in this case 1, so you would put a decimal in front of it to make it 446,220

natali 33 [55]3 years ago

4 0

Hey You!

446,221 to the nearest ten would be: 466,220

This is because 1 is less than five, so we would round it to the closest lower ten, not higher.

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What would you multiply the fraction 12/28 by to make the product equal 1

vredina [299]

28/28 is correct because one whole equals a whole pizza and so for example, there is 28 slices of pizza, which equals one BIG JUICY WHOLLLLE

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20-%202.68%20-%203.57%7D%7B%28%20-%2012.5%20%5Ctimes%200.4%29%7D%20" id="TexForm

Maslowich

Answer:

The answer is 0.125

Step-by-step explanation:

-2.68 - 3.57 ÷ (-12.5 × 4)

-6.25 ÷ (-50)

6.25 ÷ 50 = 0.125

Thus, The answer is 0.125

-TheUnknownScientist

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3 years ago

Read 2 more answers

Colin gets 3/10 of the money. Emma and dave share the rest of money in the ratio 3:2. what is dave's share of the money.

otez555 [7]

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3 years ago

4. A small high school holds its graduation ceremony in the gym. Because of seating constraints, students are limited to a maxim

Ad libitum [116K]

Answer:

(a) The mean and standard deviation of X is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of T are 390 and 180 respectively.

(c) The distribution of T is N (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.

Step-by-step explanation:

(a)

The random variable X is defined as the number of tickets requested by a randomly selected graduating student.

The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:

X P (X)

0 0.05

1 0.15

2 0.25

3 0.25

4 0.30

The formula to compute the mean is:

$\mu=\sum x\cdot P(X)$

Compute the mean number of tickets requested by a student as follows:

$\mu=\sum x\cdot P(X)\\=(0\times 0.05)+(1\times 0.15)+(2\times 0.25)+(3\times 0.25)+(4\times 0.30)\\=2.6$

The formula of standard deviation of the number of tickets requested by a student as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}\\=\sqrt{[(0^{2}\times 0.05)+(1^{2}\times 0.15)+(2^{2}\times 0.25)+(3^{2}\times 0.25)+(4^{2}\times 0.30)]-(2.6)^{2}}\\=\sqrt{1.44}\\=1.2$

Thus, the mean and standard deviation of X is 2.6 and 1.2 respectively.

(b)

The random variable T is defined as the total number of tickets requested by the 150 students graduating this year.

That is, T = 150 X

Compute the mean of T as follows:

$\mu=E(T)\\=E(150\cdot X)\\=150\times E(X)\\=150\times 2.6\\=390$

Compute the standard deviation of T as follows:

$\sigma=SD(T)\\=SD(150\cdot X)\\=\sqrt{V(150\cdot X)}\\=\sqrt{150^{2}}\times SD(X)\\=150\times 1.2\\=180$

Thus, the mean and standard deviation of T are 390 and 180 respectively.

(c)

The maximum number of seats at the gym is, 500.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Here T = total number of seats requested.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{T}=n\times \mu_{X}=390$