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2 years ago

14

Eric has two dogs. He feeds each dog 250 grams of dry food each, twice a day. If he buys a 10-kilogram bag of dry food, how many

days will the bag last

2 answers:

viktelen [127]2 years ago

8 0

It will last around 10 days

Ghella [55]2 years ago

4 0

Answer:

10

Step-by-step explanation:

okay so if he has 2 dogs and he feeds each dog 250 grams twice a day

if he buys a 10 kilo bag, how many days will it last?

so first off, convert the 250 grams into kilos. which is 0.25

0.25 × 4 = 1 kilo

4 is the amount of times he feeds the dogs.

so in reality, it should take 10 days.

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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%.

5 0

2 years ago

A certain shade of green paint is made from 5 parts yellow mixed with three parts blue if 2 cans of yellow are used how many can

wariber [46]

Because the ratio of yellow to blue can be expressed as 5/3, you would solve the equation 5/3= 2/x. To solve, you would get x alone by cross multiplying and getting 5x=6. Divide both sides by 5 and get 6/5 or 1 and 1/5 cans of blue paint

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2 years ago

Please help fast!!!!!

OverLord2011 [107]

Answer:

60 i think

Step-by-step explanation:

7 0

2 years ago

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A car traveled 23 miles in 15 minutes. how many miles per hour was it traveling?

NNADVOKAT [17]

Answer:

1.533333 miles per hour

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2 years ago

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In the study of population dynamics one of the most famous models for a growing but bounded population is the logistic equation

MA_775_DIABLO [31]

Answer:

If $K$ is a constant of integration, then

$P = {\displaystyle \frac{1}{b/a + Ke^{-at}}}$

Step-by-step explanation:

According to the information of the problem we know that

$\frac{dP}{dt} = P(a-bP)$

Remember that in general a Bernoulli equation is an equation of the type

$y' + p(x)y = q(x)y^n$

And the idea to solve the equation is to substitute

$v = y^{1-n}$

Now for this case

$\frac{dP}{dt} - Pa = -bP^2$

Then we substitute

$v = P^{1-2} = P^{-1}$

Therefore

$P = v^{-1}$

and if you compute the derivative of that you get that

$\frac{dP}{dt} = -v^{-2} \frac{dv}{dt}$

Now you substitute that onto the original equation and get

$\frac{dP}{dt} - Pa = -bP^2$

$-v^{-2} \frac{dv}{dt} - v^{-1} = -bv^{-2$

If you multiply everything by $-v^2$ you get that

$\frac{dv}{dt} + v = b$

That's a linear differential equation and the solution would be

$v = {\displaystyle \frac{b}{a} + Ke^{-at}} = P^{-1}$

Where $K$ is a constant of integration, then

$P = {\displaystyle \frac{1}{b/a + Ke^{-at}}}$

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2 years ago