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3 years ago

The following is an example of what property?

2 answers:

6 0

This is known as an Addition Identity Property.

It's a rule that states that if you add zero to a number, you would get the same same number as the answer.

Example: 10 + 0 = 10

So the answer is D.

NISA [10]3 years ago

4 0

If you solve it, you get x=0. That means you are adding zero, so you are dealing with the additive identity property.

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X^3-6x^2-7x+60=0 i need help

alexgriva [62]

Answer:

x= -3, 4 or 5

Step-by-step explanation:

if you factor out the expressions, they look like this:

(x+3)(x-4)(x-5) = 0

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3 years ago

Can anyone help me with this answer -8(-5w+y-2)

Tatiana [17]

40w - 8y + 16
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3 years ago

What is 12 divided by 8

Alina [70]

Answer:

1.5

Step-by-step explanation:

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3 years ago

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Suppose 10 people are chosen at random from a group of 1200. What is the probability that all 10 would rate their financial shap

Pachacha [2.7K]

1200/10= 120 and 120 is your answer

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4 years ago

Suppose that prices of a certain model of a new home are normally distributed with a mean of $150,000. Use the 68-95-99.7 rule t

levacccp [35]

Answer:

68% of buyers paid between $147,700 and $152,300.

Step-by-step explanation:

We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.

Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.

Let X = prices of a certain model of a new home

SO, X ~ Normal( $\mu=150,000 ,\sigma=2,300$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean price = $150,000

$\sigma$ = standard deviation = $2,300

Now, according to 68-95-99.7 rule;

Around 68% of the values in a normal distribution lies between $\mu-\sigma$ and $\mu-\sigma$ .

Around 95% of the values occur between $\mu-2\sigma$ and $\mu+2\sigma$ .

Around 99.7% of the values occur between $\mu-3\sigma$ and $\mu+3\sigma$ .

So, firstly we will find the z scores for both the values given;

Z = $\frac{X-\mu}{\sigma}$ = $\frac{147,700-150,000}{2,300}$ = -1

Z = $\frac{X-\mu}{\sigma}$ = $\frac{152,300-150,000}{2,300}$ = 1

This indicates that we are in the category of between $\mu-\sigma$ and $\mu-\sigma$ .

SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.

7 0

4 years ago