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aleksley [76]
3 years ago
6

Ken has 4.66 pounds of walnuts ,2.1 pounds of cashew , and 8 pounds of peanuts .he mixes them together and divides them equally

among 18 bags .how many pounds of nuts are in each bag ?
Mathematics
1 answer:
kogti [31]3 years ago
7 0
<span>walnut= 4.66 pounds cashew=2.1 pounds peanuts=8 pounds Total nuts= 4.66+2.1+8 = 14.76 pounds each bag contains= 14.76/18=0.82 pounds of nuts</span>
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If 2 pounds of bananas cost $1.89, what is the cost per pound? Show your work
OLEGan [10]
You can set up an expression where p = the pounds of bananas.
With that information, 2p =$1.89
To figure out the cost per pound, you want the cost of 1 p. Therefore, you divide both sides by 2. 
p = $1.89/2
p = $0.945
(since it is money you normally round to two decimal places so you would say $0.95 or 95 cents)
6 0
3 years ago
A square in the board below and a letter in the word ATTENTION is chosen at
kolezko [41]

Answer:

4 /15

Step-by-step explanation:

From the number cube :

Count of Numbers greater than 4 = 16

Sample space of number cube = 20

P(number greater than 4) = 16 / 20 = 4/5

Word : ATTENTION

Count of letters in attention = 9

Count of T's in attention = 3

P(choosing T) = 3 / 9 = 1/3

P(greater than 4, then T) = 4/5 * 1/3 = 4 /15

5 0
2 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

And the margin of error is given by:

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

±0.75 ounce

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