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3 years ago

12

A sports store sells a 4 pack of tennis balls for $11.49, a 6 pack of tennis balls for $16.70, and a 9 pack for $22.99. Which pa

ck is the best buy? Explain

2 answers:

garik1379 [7]3 years ago

4 0

Answer:

9 pack

Step-by-step explanation:

11.49/4=$2.87 per ball

16.70/6=$2.78 per ball

22.99/9=$2.55 per ball

Luba_88 [7]3 years ago

3 0

The pack of 4 is better because you don’t have to spend as much money as the 6 and 9 packs

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olga_2 [115]

Answer:

a

The probability that the selected joint was judged to be defective by neither of the two inspectors is $P(A' n B' ) = 0.8855$

b

The probability that the selected joint was judged to be defective by inspector B but not by inspector A is $P(A' n B) =0.0403$

Step-by-step explanation:

From the question we are told that

The sample size is $n_s = 10000$

The number of outcome for inspector A is $n__{A}} = 742$

The number of outcome for inspector B is $n__{B}} = 745$

The number of joints judged defective by both inspector is $n(A u B) = 1145$

The the probability that the selected joint was judged to be defective by neither of the two inspectors is mathematically represented as

$P(A' n B' ) = 1 - P(A u B)$

Now

$P(A\ u \ B) = \frac{n(Au B)}{n_s }$

substituting values

$P(A\ u \ B) = \frac{1145}{ 10 000 }$

So

$P(A' n B' ) = 1 - \frac{1145}{10 000}$

$P(A' n B' ) = 0.8855$

the probability that the selected joint was judged to be defective by inspector B but not by inspector A is mathematically represented as

$P(A' n B) = P(A \ u \ B) -P(A)$

Now

$P(A) = \frac{n__{A}}{n_s}$

substituting values

$P(A) = \frac{742}{10 000}$

So

$P(A' n B) = \frac{1145}{10 000} - \frac{742}{10 000}$

$P(A' n B) =0.0403$

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Given right triangle ABCABC with altitude \overline{BD}

Alex

Answer:

$x = 2\sqrt{11}$

Step-by-step explanation:

AD = 4

AC = 11

DC = 11 - 4 = 7

First, find BD using the right triangle altitude theorem:

$BD = \sqrt{AD*DC}$

Plug in the values

$BD = \sqrt{4*7}$

$BD = 2\sqrt{7}$

Use pythagorean theorem to find x:

x² = AD² + BD²

Plug in the values

$x^2 = 4^2 + (2\sqrt{7})^2$

$x^2 = 4^2 + (2\sqrt{7})^2$

$x^2 = 16 + (4*7)$

$x^2 = 16 + 28$

$x^2 = 44$

Take the square root of both sides

$\sqrt{x^2} = \sqrt{44}$

$x = \sqrt{44}$

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$x = 2\sqrt{11}$

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10-7x=2-8x

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