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2 years ago

"A bottle of olive oil contains 1.4 qt of olive oil. What is that volume in milliliters?"

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

3 0

Answer:

The volume of oil in milliter is 1,324.5 .

Step-by-step explanation:

Volume of olive oil = 1.4 qt = 1.4 Quarts

And we know that 1 Liter is equal to the 1.057 Quarts.

1 L = 1.057 Quarts

Then 1.4 Quarts will be equal to:

$1.4 Quarts=\frac{1.4}{1.057} L=1.3245 L$

1 Liter is equal to the 1000 milliliter.

1 L = 1000 mL

$1.3245 L=1.3245\times 1000=1,324.5 mL$

The volume of oil in milliter is 1,324.5 .

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suter [353]

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n=
\begin{array}{llll}
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1 year ago

Read 2 more answers

Determine the product: (46.2 × 10–1) ⋅ (5.7 × 10–6). Write your answer in scientific notation.

lianna [129]

Answer:

Step-by-step explanation:

10-6 X 10-1 = 10-7

5.7*46.2=263.34

263.34=2.6334 x 10^2

10^2 x 10^-7 = 10^-5

=2.6334 x 10-5

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1 year ago

Harvey has “r” red peppers and one fourth as many orange peppers. Choose the expression that shows how many orange peppers Harve

lana [24]

Answer:

$\frac{1}{4}r$

Step-by-step explanation:

The amount of red pepper that Harvey has is r red peppers and one fourth as many orange peppers. What this means is that the number of orange pepper that Harvey has is one fourth the number of red peppers. Since the number of red pepper is r, therefore the number of orange pepper is given as:

Number of orange pepper = $\frac{1}{4} \ of\ red\ pepper(r)$

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1 year 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

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

So, 90% confidence interval for the population proportion, p is ;

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

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

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

= [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 = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$