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4 years ago

Addition properties and subraction rules

1 answer:

3 0

You don't have to add in a certain order it's any way but when your adding domt forget to cary if you need to. For subtraction you have to the way it gives you and you have to borrow when you can subtract a number that won't go into. For an example 90-12

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What is the explicit rule for the sequence?

Triss [41]

Answer: D

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Round to the nearest tenth.

Minchanka [31]

From the formula given:

A = 150 ft², π ≈ 3.14

r = √(A/π)

r = √(150/π)

r = √(150/3.14) use calculator

r ≈ 6.911

r ≈ 6.9

The tenth digit is 9, and the digit after it is 1, and that is not up to 5, so the 9 is not rounded up to 10, it remains as 9.

Radius is ≈ 6.9 feet to nearest tenth

Hope this helps.

4 0

3 years ago

Suppose that a basketball player can score on a particular shot with probability .3. Use the central limit theorem to find the a

Rom4ik [11]

Answer:

(a) The probability that the number of successes is at most 5 is 0.1379.

(b) The probability that the number of successes is at most 5 is 0.1379.

(d) The probability that the number of successes is at most 11 is 0.9357.

→ All the exact probabilities are more than the approximated probability.

Step-by-step explanation:

Let S = a basketball player scores a shot.

The probability that a basketball player scores a shot is, P (S) = p = 0.30.

The number of sample selected is, n = 25.

The random variable $S\sim Bin(25,0.30)$

According to the central limit theorem if the sample taken from an unknown population is large then the sampling distribution of the sample proportion ( $\hat p$ ) follows a normal distribution.

The mean of the the sampling distribution of the sample proportion is: $E(\hat p)=p=0.30$

The standard deviation of the the sampling distribution of the sample proportion is:

$SD(\hat p)=\sqrt{\frac{ p(1- p)}{n} }=\sqrt{\frac{ 0.30(1-0.30)}{25} }=0.092$

(a)

Compute the probability that the number of successes is at most 5 as follows:

The probability of 5 successes is: $p=\frac{5}{25} =0.20$

$P(\hat p\leq 0.20)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.20-0.30}{0.092} )\\=P(Z\leq -1.087)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 5 is 0.1379.

The exact probability that the number of successes is at most 5 is:

$P(S\leq 5)={25\choose 5}(0.30)^{5}91-0.30)^{25-5}=0.1935$

The exact probability is more than the approximated probability.

(b)

Compute the probability that the number of successes is at most 7 as follows:

The probability of 5 successes is: $p=\frac{7}{25} =0.28$

$P(\hat p\leq 0.28)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.28-0.30}{0.092} )\\=P(Z\leq -0.2174)\\=1-P(Z$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 7 is 0.4129.

The exact probability that the number of successes is at most 7 is:

$P(S\leq 57)={25\choose 7}(0.30)^{7}91-0.30)^{25-7}=0.5118$

The exact probability is more than the approximated probability.

(c)

Compute the probability that the number of successes is at most 9 as follows:

The probability of 5 successes is: $p=\frac{9}{25} =0.36$

$P(\hat p\leq 0.36)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.36-0.30}{0.092} )\\=P(Z\leq 0.6522)\\=0.7422$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 9 is 0.7422.

The exact probability that the number of successes is at most 9 is:

$P(S\leq 9)={25\choose 9}(0.30)^{9}91-0.30)^{25-9}=0.8106$

The exact probability is more than the approximated probability.

(d)

Compute the probability that the number of successes is at most 11 as follows:

The probability of 5 successes is: $p=\frac{11}{25} =0.44$

$P(\hat p\leq 0.44)=P(\frac{\hat p-E(\hat p)}{SD(\hat p)}\leq \frac{0.44-0.30}{0.092} )\\=P(Z\leq 1.522)\\=0.9357$

**Use the standard normal table for probability.

Thus, the probability that the number of successes is at most 11 is 0.9357.

The exact probability that the number of successes is at most 11 is:

$P(S\leq 11)={25\choose 11}(0.30)^{11}91-0.30)^{25-11}=0.9558$

The exact probability is more than the approximated probability.

6 0

4 years ago

Three toy reindeer and two boxes of cookies cost $13.00.Two toy reindeer and two boxes of cookies cost $10.00. What is the unit

Semmy [17]

Answer:

The unit cost of one toy reindeer is $3 and the unit cost of one box of cookies is $2

Step-by-step explanation:

Let

x ----> the unit cost of one toy reindeer

y ----> the unit cost of one box of cookies

we know that

$3x+2y=13.00$ ----> equation A

$2x+2y=10.00$ ----> equation B

Solve the system by graphing

The solution is the intersection point both graphs

using a graphing tool

The solution is the point (3,2)

see the attached figure

therefore

The unit cost of one toy reindeer is $3 and the unit cost of one box of cookies is $2

4 0

4 years ago

What's the equivalent to the expression of (3×-8)

Zarrin [17]

That expression is equivalent to 1/2(6-16) because since you are distributing 1/2 you are going to divide 6 and -16 by 1/2 and that will give you 3x-8

5 0

4 years ago