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3 years ago

Can someone help ASAP (8 grade math)

Mathematics

2 answers:

GrogVix [38]3 years ago

8 0

Answer:

H) 62,500

Step-by-step explanation:

6x3% ( from the 3% because its annual and it says for 6 years )=.18

53,000x.18=9,540

53,000+9,540=62,540

62,540 is closest to H)62,500

gladu [14]3 years ago

4 0

Answer: $9,500

Step-by-step explanation:

It asks how much in interest not the total value

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Solve for x: 3(x + 1) = −2(x − 1) − 4.

11Alexandr11 [23.1K]

Its x=−1 here the answer

7 0

3 years ago

Read 2 more answers

Why do you flip the inequality sign when dividing by a negative number?

ohaa [14]

Assume x is 100.

-x<1
-100 is smaller than 1.

Dividing by -1,
x>-1
100 is greater than -1
If the inequality is not changed,
100<-1
100 is less than -1 which is not true

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4 years ago

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%7C3%20-%20x%7C%20%20-%20%28x%20-%201%29%20%3D%204" id="TexFormula1" title=" |3 - x| - (x

aleksley [76]

Answer:

$x=0$

Step-by-step explanation:

Evaluate all the arithmetic first, and then remember absolute value equations have two possible ways to go (+ and -).

You can see how I split them on the 3rd line, later demonstrating how the 2nd one is invalid.

$|3-x|-x+1=4\\\\|3-x|-x=3\\\\3-x-x=3\\-(3-x)-x=3\\\\3-2x=3\\-3+x-x=3\\\\-2x=0\\-3\neq 3\\\\x=0\\-3\neq 3\\$

Therefore, the solution is $x=0$

4 0

3 years ago

Tony rounded each of the numbers 1, 143 and 1, 149 to the nearest hundred which word correctly compares the rounding numbers

Montano1993 [528]

Full Question:

Tony rounded each of the numbers 1,143 and 1,149 to the nearest hundred. Which choice correctly compares the rounded numbers?

$1,000 = 1,000$

$1,140< 1,150$

$1,100 = 1.100$

$1,140>1,150$

Answer:

$1,100 = 1,100$

Step-by-step explanation:

Given

1,143 and 1,149

Required

Which of the option is correct

We start by approximating both numbers to nearest digit

1,143; when approximated to nearest hundred is 1,100

1,149; when approximated to nearest hundred is also 1,100

Hence;

1,143 ≅ 1,100

1,149 ≅ 1,100

Comparing both results, we have that

$1,100 = 1,100$

From the list of given options, option C is correct;

4 0

4 years ago