What is 119,000,003 in expanded form

Pavel [41]

Step-by-step explanation:

I have no idea what you are asking to do lol

8 0

1 year ago

Tarush is a landscape architect. For his first public project he is asked to create a small scale drawing of a garden to be plac

Lunna [17]

Answer:

See the attached figure 2.

Step-by-step explanation:

The question is as shown at the attached figure 1.

given: The garden is a right triangle with base 10 m and height 15 m.

We need to draw the garden such that 1 unit on the grid represents 2 m.

So the scale factor is 1 unit = 2 m

base = 10 m = 10/2 = 5 units

height = 15 m = 15/2 = 7.5 units

So, the garden on the grid will be a right triangle with base 5 units and height 7.5 units.

See the attached figure.

3 0

1 year ago

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Stores often mark up items each year to cover inflation. Last year, a CD at Juan’s Tunes cost $12.95. This year, the cost was ma

Ratling [72]

4*12.95*1.08-4*12.95*1.08*.10
if u buy 4

7 0

1 year ago

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Find the balance in the account. $800 principal earning 7%, compounded annually, after 4 years

scoray [572]

800×(1.07)^(4) =1,048.63

6 0

2 years ago

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g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago