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igor_vitrenko [27]
3 years ago
15

Samuel is going to a carnival that has games and rides. Each game costs $2 and each ride costs $5. Samuel spent $60 altogether o

n 18 games and rides. Determine the number of games Samuel played and the number of rides Samuel went on.
Samuel played__games and went on__rides rides.

Pleases show how you worked it out!
Mathematics
1 answer:
Stells [14]3 years ago
6 0
Think of it as a problem with only the equations so we know the exact numbers of each ride and each cost now we need to do some division and multiplication
2 games and 4 rides
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What is the vertex of function f(x)=(x-3)(x+5)?
butalik [34]

Answer:

( − 1 , − 16 )

Step-by-step explanation:

6 0
2 years ago
Bethany paid $408.75 for a pair of diamond stud earrings. The sale price was $375. What is the Percent of Sales Tax she paid for
9966 [12]

Answer:

She paid 9% of sales tax

Step-by-step explanation:

The sale price of diamond = $375

Let x be percent of sales tax

So, amount of tax = x\% \times 375 = \frac{x}{100} \times 375 =3.75x

So, Amount including tax = 375+3.75x

We are given that Bethany paid $408.75 for a pair of diamond stud earrings.

So, 375+3.75x=408.75

3.75x=408.75-375

x=\frac{408.75-375}{3.75}

x=9

Hence She paid 9% of sales tax

5 0
3 years ago
In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
A square is always an example of a: I. rectangle II. rhombus III. parallelogram I I, II, and III none of these II
Igoryamba
It's always a Parallelogram
4 0
3 years ago
X + 5y - 10 = 0<br> x = 2y - 8
ipn [44]
So since we know what x is, we can substitute it into the original equation for x like so to solve for y...

(2y - 8) + 5y - 10 = 0
2y - 8 + 5y = 10
2y + 5y = 18
7y = 18
y = 18/7 (or about 2.57)

So now we know what x is, we can sub it into the below equation to solve for x...

x = 2(18/7) - 8
x = 36/14 - 8 (or about -5.43)


3 0
3 years ago
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