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Naddika [18.5K]

Answer: See explanation (btw, the answer is always in bold letters)

Step-by-step explanation:

1.) 1 gallon for $1.99

1/2 gallon for $0.98

1 gallon would be 0.98 × 2 = <em>1.96</em>

2 gallons for $3.98

1 gallon would be 3.98 ÷ 2 = 1.99

2.) 621 ÷ 4 = 155.25

He pays $155.25 each payment!

3.) V = s³

s = 3

V = 3³

3 × 3 × 3 = 27

V = 27 m³

4.) This is a rectangle; the area of a rectangle is l × b.

L = 6.9

l × b = a

6.9b = 28.29

Divide both sides by 6.9

b = 4.1

The missing length is 4.1 m

5.) False

5 0

3 years ago

A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters

PilotLPTM [1.2K]

Answer:

e. 0.0072

Step-by-step explanation:

We are given that a bottling company uses a filling machine to fill plastic bottles with cola. And the contents vary according to a Normal distribution with Mean, μ = 298 ml and Standard deviation, σ = 3 ml .

Let Z = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1) where, Xbar = mean contents of six randomly

selected bottles

n = sample size i.e. 6

So, Probability that the mean contents of six randomly selected bottles is less than 295 ml is given by, P(Xbar < 295)

P(Xbar < 295) = P( $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{295 - 298}{\frac{3}{\sqrt{6} } }$ ) = P(Z < -2.45) = P(Z > 2.45)

Now, using z% score table we find that P(Z > 2.45) = 0.00715 ≈ 0.0072 .

Therefore, option e is correct .

7 0

3 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$