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jeyben [28]
3 years ago
6

Could someone pls help:(

Mathematics
2 answers:
dsp733 years ago
7 0

Answer:

4.5

Step-by-step explanation:

3^2=9

1/2*9=4.5

jekas [21]3 years ago
4 0

Answer:

g(2) = 9/2

Step-by-step explanation:

You would plug in 2 into g(x):

g(2) = 1/2 (3)2

3 to the power of 2 is 9;

g(2) = 1/2 * (9)

multiply:

g(2) = 9/2

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What is the difference in the costs for a one-night stay with and without breakfast?
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Answer:

$23.12

Step-by-step explanation:

With breakfast included the total would be $142.99. Without breakfast it would be $119.87. So to get the difference it would be:

$142.99 - $119.87 = <em>$23.12</em>

6 0
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D square=3h+s<br>change of subject formula<br>make h the leader<br>​
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Answer: h=(D²-s)/3

Step-by-step explanation: D² =3h+s

Subtract a from both sides of the equation

D²-s =3h+s-s

Simplifying,

D²-s =3h

Divide both sides by 3( to isolate 'h')

(D²-s)/3 =3h/3

Simplifying,

(D²-s)/3 =h

Hence, h= (D²-s)/3

Hope it helps!

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2 years ago
Lance rolls a die 5 times. What is the probability that he rolls an even number all five times?
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Read 2 more answers
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

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