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3 years ago

A recipe for lemonade calls for 1 cup of sugar and 6 cups of water. How much sugar is used per cup of water?

Mathematics

1 answer:

grandymaker [24]3 years ago

3 0

The answer is 1/6, or .16666...

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Find the domain and range of the relation.<br> {(3,-2) (2,1) (-1,0) (1,3)}

svp [43]

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

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2 years ago

Help pls Use the substitution method to solve the system of equations. Choose the

AURORKA [14]

If I’m using substitution the answer is A

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3 years ago

Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and

Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437$

P(sample average sediment density is at most 3.00)

$P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)$

Calculation the value from standard normal z table, we have,

$P(x \leq 3.00) = 0.918$

0.918 is the probability that the sample average sediment density is at most 3.00

4 0

3 years ago

Determine the area of the parallelogram below

spayn [35]

Answer:

A= BH

A=8cm ×2.2 cm

A= 17.6cm²

8 0

3 years ago

An article reported that for a sample of 55 kitchens with gas cooking appliances monitored during a one-week period, the sample

Sati [7]

Answer:

Step-by-step explanation:

Given that

sample size n = 55: x bar = 654.16 and s = sample sd = 162.34

Std error = 162.34/sqrt 55 = 21.889

For 95% CI we can use t critical value as population std dev is not known.

df = 54

t critical = 2.004

Margin of error = 2.004 *21.889 = 43.866

Confidence interval lower bound = 654.16-43.866 =610.294

Upper bound = 654.16+43.866=698.026

Confidence interval rounded off at 95% = (610.29, 698.23)

8 0

3 years ago