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4 years ago

11

What is 3a^2×c^-2÷4b^0 ??? I need help

1 answer:

Lyrx [107]4 years ago

8 0

I hope this picture helps

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20 is 60% of what number

Lilit [14]

Answer:

33.33

Step-by-step explanation:

60% of 33.33 is 20. 100% of 33.33 is 33.33, therefore 60 percent of 33.33 equals 20.

3 0

3 years ago

Read 2 more answers

Evan lives in Stormwind City and works as an engineer in the city of ironforge in the morning he has three Transportation option

denis-greek [22]

Answer:

The probability that he teleports at least once a day = $\mathbf{\frac{5}{9}}$

Step-by-step explanation:

Given -

Evan lives in Stormwind City and works as an engineer in the city of ironforge in the morning he has three Transportation options teleport ride a dragon or walk to work and in the evening he has the same three choices for his trip home.

Total no of outcomes = 3

P( He not choose teleport in the morning ) = $\frac{2}{3}$

P( He not choose teleport in the evening ) = $\frac{2}{3}$

P ( he choose teleports at least once a day ) = 1 - P ( he not choose teleports in a day )

= 1 - P( He not choose teleport in the morning ) $\times$ P( He not choose teleport in the evening )

= $1 - \frac{2}{3}\times\frac{2}{3}$

= $\frac{5}{9}$

3 0

3 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.

<u><em>Let X = prices of a certain model of a new home</em></u>

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

<u>Now, according to 68-95-99.7 rule;</u>

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

A intersection b intersection c compliment

faust18 [17]

B intersection is the correct answer

7 0

3 years ago

Read 2 more answers

The heat flow vector field for conducting objects is F = - k T. where T(x, y, z) is the temperature in the object and k is a con

wolverine [178]

Answer:

The answer has been given in the attachment

Step-by-step explanation:

8 0

3 years ago