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3 years ago

How to solve 4x^4-6x^3+7/-4x^4

Mathematics

2 answers:

podryga [215]3 years ago

7 0

Answer:

9x^4/4 - 6x^3

Step-by-step explanation:

Group like terms

Add similar elements

Factor out common terms

Multiply.

**Detailed steps are in the picture attached**

I hope this helped!

Brainliest is always appreciated :)

stepladder [879]3 years ago

7 0

256x-216x-0.03x
=39.97x

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Answer:

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Step-by-step explanation:

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3 years ago

What is a pendulum?

dezoksy [38]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

4(x+9)=8x-7 Solve for x

valentina_108 [34]

Answer:

x = 10.75

Step-by-step explanation:

4(x+9)=8x-7

Step 1: Distribute the 4 to the x and 9

4 * x = 4x

4 * 9 = 36

We now have 4x + 36 = 8x - 7

Step 2: add 7 to both sides

36 + 7 = 43

-7 + 7 cancels out

We now have 4x + 43 = 8x

Step 3 subtract 4x from both sides

4x - 4x cancels out

8x - 4x = 4x

We now have 4x = 43

step 4 divide both sides by 4

4x / 4 ( the 4s cancel out and we're left with x )

43/4 = 10.75

We're left with x = 10.75

3 0

2 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

3 years ago

Find the simple interest owed for the use of the money. Assume 360 days in a year. P=16,500, r=6%, t=180 days

Taya2010 [7]

The formula for simple interest is the following:

$S=Prt$

In this formula, $S$ is the amount of interest, $P$ is the principal (or initial) amount, $r$ is the interest rate, and $t$ is the length f time, typically in years.

First, let's make a couple of changes to how we wrote down the variables.

Change the interest rate, $6\%$ , to the decimal value $0.06$ .
Convert the length of time to years. 180 days divided by 360 days per year equals $0.5$ years.

Substitute in the values and solve with a calculator.

$\begin{aligned}S&=16,500\cdot0.06\cdot0.5\\&=\boxed{495}\end{aligned}$

7 0

1 year ago