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1 year ago

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At the football game they sold $4 pizzas and $2 sodas, which made the school $260. The number of sodas sold was 5 more than thre

e times the number of pizzas sold. Determine the amount of pizza and sodas sold.

1 answer:

Firlakuza [10]1 year ago

3 0

25 Pizzas and 80 Sodas

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The Hernandez family orders 3 large pizzas. They cut the pizzas so that each pizza has the same number of slices, giving them a

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Answer:

<h2>c</h2>

Step-by-step explanation:

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2 years ago

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Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of

Gelneren [198K]

Answer:

a

$P(X < 24 )= 21.186\%$

b

$x = 19.78 \ months$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 26 \ months$

The standard deviation is $\sigma = 4 \ months$

Generally 2 year is equal to 24 months

Generally the percentage of total production will the company expect to replace is mathematically represented as

$P(X < 24 )= P(\frac{X - \mu }{ \sigma} < \frac{24 - 26}{4} )$

Generally $\frac{X - \mu}{\sigma } =Z (The \ standardized \ value \ of \ X )$

$P(X < 24 )= P(Z < -0.8 )$

Generally from the z-table

$P(Z < -0.8) = 0.21186$

So

$P(X < 24 )= 0.21186$

Converting to percentage

$P(X < 24 )= 0.21186 * 100$

=> $P(X < 24 )= 21.186\%$

Generally the duration that should be the guarantee period if Accrotime does not want to make refunds on more than 6% is mathematically evaluated as

$P(X < x) = P(\frac{X - \mu }{\sigma} < \frac{x - 26}{4} )= 0.06$

=> $P(X < x) = P(Z < \frac{x - 26}{4} )= 0.06$

From the normal distribution table the z-score for 0.06 at the lower tail is

$z = -1.555$

So

$\frac{x - 26}{4} = -1.555$

=> $x = 19.78 \ months$

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1 year ago

Suppose that you want to mix two coffees in order to obtain 100 pounds of a blend. If x represents the number of pounds of coffe

Svetach [21]

Answer: $100-x$

Therefore, the algebraic exp

Step-by-step explanation:

Given : x represents the number of pounds of coffee A.

The total weight of the mix of coffee A and coffee B = 100 pounds.

Then , we have the following expression to represents the number of pounds of coffee B:-

$100-x$

Therefore, the algebraic expression that represents the number of pounds of coffee B. :-

$100-x$

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2 years ago

A 0.143-Henry Inductor is connected in series with a variable resistor to a 208-volt 400-cycle source. For what value of capacit

aleksandr82 [10.1K]

Answer:

A.)359.2, B.)2.5 uf

Step-by-step explanation:

E / I = R

208 / 1.04 = 200 ohms

2*pi*f*L = Xl

6.28*400*.143 = 359.2 ohm

1 / (2*pi*f*Xc) = c

1 /(6.28*400*159.2) = 2.5 uf

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2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

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

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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2 years ago