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maxonik [38]
3 years ago
10

You had $20 to spend on seven avocados. After buying them you had $6. How much did each avocado cost?

Mathematics
2 answers:
mestny [16]3 years ago
8 0
They each cost 2 dollars

Fantom [35]3 years ago
5 0
$20 - $6 = $14 so she spent $14 on avocados. if she bought 7 of them, each avocado would equal the same amount so 14 ÷ 7 = 2

so she spent $2 on each avocado.
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Find the product of 52.6 x 9
marta [7]

Answer: 473.4

Step-by-step explanation: if you can't solve it, break the first number apart, multiply both numbers, then add them together. That way there not such big numbers :)

8 0
3 years ago
Evaluate the expression 11C3
butalik [34]
Evaluating the expression 11C3 is very simple to do. 

If you are assuming that C is the notation for the combination. 
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3 0
3 years ago
Kevin and Randy Muise have a jar containing 84 coins, all of which are either quarters or nickels. The total value of the coins
ella [17]
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4 0
3 years ago
In preparation for an earnings report, a large retailer wants to estimate p= the proportion of annual sales
mr Goodwill [35]

Using the z-distribution, it is found that the 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

The sample size and the estimate are given by:

n = 161, \pi = \frac{37}{161} = 0.2298

Hence:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 - 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.1648

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2298 + 1.96\sqrt{\frac{0.2298(0.7702)}{161}} = 0.2948

The 95% confidence interval for the proportion of sales that occured in December is (0.1648, 0.2948).

More can be learned about the z-distribution at brainly.com/question/25890103

5 0
2 years ago
PLZZZZ NEED HELP!!!!!!
jeka94
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