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2 years ago

Simplify each equation. Show steps. 1. 2(x + 3) + 4x 2. 4 + 2(x – 2) – 4 3. 6(2x – 4) – 3x 4. -6x – 4 + 5x + 7

Mathematics

2 answers:

Lelu [443]2 years ago

7 0

Answer:

Step-by-step explanation:

1. 2x + 6 + 4x = 6x + 6

2. 4 + 2x - 4 - 4 = 2x - 4

3. 12x - 24 - 3x = 9x - 24

4. -x + 3

Margarita [4]2 years ago

4 0

1) 9x-24
2)2x+4
3) 9x-24
4)-x+3

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ILL GIVE BRAINLY THING

Vinil7 [7]

Answer:

the slope is -4/3

Step-by-step explanation:

you can either use the graph or algebra to find the slope.

to use the graph, simply count how many spaces up/down and to the left/right that you go.

to use algebra, take the coordinates of your two points and do (y2-y1)/(x2-x1).

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2 years ago

Find the value of b that will make the equation sin 62°= cos B true when 0°

KiRa [710]

Hello from MrBillDoesMath!

Answer:

b = 28 degrees

Discussion:

In general,

sin (90 - @) = cos(@)

Set 90 [email protected] = 62 => @ = 28 so

sin (90-28) = sin(62) = cos(28) => b = 28

Thank you,

MrB

4 0

2 years ago

A $3,500 loan, taken now, with a simple interest rate of 5% per year, will require a total repayment of $3,850. At what time t w

Alecsey [184]

Answer:

Roughly 2 years

Step-by-step explanation:

3500(1.05)^t=3850

(1.05)^t=1.1

t=1.953

So roughly 2 years

5 0

2 years ago

Read 2 more answers

If $12000 is invested in an account in which the interest earned is continuously compounded at a rate of 2.5%

vladimir1956 [14]

Complete Question

If $12000 is invested in an account in which the interest earned is continuously compounded at a rate of 2.5% for 3 years

Answer:

$ 12,934.61

Step-by-step explanation:

The formula for Compound Interest Compounded continuously is given as:

A = Pe^rt

A = Amount after t years

r = Interest rate = 2.5%

t = Time after t years = 3

P = Principal = Initial amount invested = $12,000

First, convert R percent to r a decimal

r = R/100

r = 2.5%/100

r = 0.025 per year,

Then, solve our equation for A

A = Pe^rt

A = 12,000 × e^(0.025 × 3)

A = $ 12,934.61

The total amount from compound interest on an original principal of $12,000.00 at a rate of 2.5% per year compounded continuously over 3 years is $ 12,934.61.

7 0

2 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

$Z = \frac{X - \mu}{s}$

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

$Z = \frac{X - \mu}{s}$

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago