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2 years ago

5

. Laser tag costs $35 per game. The laser tag fans want to play more than 1 game of laser tag. Write an expression to help you f

ind the cost of playing more than 1 game of laser tag. (2 points: 1 point for defining the variable, 1 point for the expression)

1 answer:

8_murik_8 [283]2 years ago

3 0

Let x represent the amount of games played.

When x is used, it need to be multiplied by the price ($35).

x(35) = cost in $

Hope this helped :)

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A number increased by 8 gives the same result as a number multiplied by 8. whay is that number?

Airida [17]

$x+8=8x\\
7x=8\\
x=\frac{8}{7}$

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3 years ago

Cameron needs to order some new supplies for the restaurant where he works. The

zvonat [6]

Answer:

598=194+8x

Step-by-step explanation:

To solve you would subtract 194 from its self and 598

598=194+8x

-194 -194

404=8x

then you would divide 8 from it's self and 404

which would make it 50.5

(but you cant buy half a package of spoons so realistically it'd be 50 or 51.

hope this helped :)

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2 years ago

Read 2 more answers

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago

What is the area of the following trapezoid

nignag [31]

Answer:

d. 130

Step-by-step explanation:

3 0

2 years ago

Need answer to 4 and 5

algol13

Answer:

Substitute n =1, 2, 3, 4, 5 to get the first five terms.

Step-by-step explanation:

(4). $g_n = 2 . 3^{n - 1}$

To find the first term, substitute n = 1.

Therefore, $g_1 = 2 . 3^{1 - 1} = 2 . 3^{0} = 2 . 1$ = 2

$g_2 = 2 . 3^{2 - 1} = 2 . 3$ = 6

$g_3 = 2 . 3^{3 - 1} = 2 . 3^{2} = 2 . 9$ = 18

$g_4 = 2 . 3^{4 - 1} = 2 . 3^{3} = 2 . 27$ = 54

$g_5 = 2 . 3^{5 - 1} = 2 . 3^{4} = 2 . 81$ = 168

Therefore the first five terms of the sequence are: 2, 6, 18, 54, 168.

(5). $t_n = \frac{2}{3}t_{n - 1}$

$t_2 = \frac{2}{3} t_{2 - 1} = \frac{2}{3}t_1 = \frac{2}{3}6$ = 4

$t_3 = \frac{2}{3} t_2 = \frac{2}{3}. 4 = \frac{8}{3}$

$t_4 = \frac{2}{3}t_3 = \frac{2}{3}\frac{8}{3} = \frac{16}{9}$

$t_5 = \frac{2}{3}t_4 = \frac{2}{3}\frac{16}{9} = \frac{32}{27}$

Therefore the first five terms in the sequence are: 6, 4, 8/3, 16/9, 32/27.

8 0

3 years ago