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3 years ago

Hockey practice begins at 6:17 p.m.

Mathematics

1 answer:

Ket [755]3 years ago

3 0

Answer:

51 minutes ???????????

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(2 p + 7)(3 p - 9)

scZoUnD [109]

$(2p+7)(3p-9)\\6p^2-18p+21p-63\\6p^2+3p-63$

Your answer is 2. $6p^2+3p-63$ !

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7 0

3 years ago

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What does s(8) = 6.0 mean in terms of this problem?

Olegator [25]

Hi I think this answer would be b. A is backwards. C and d do not make sense

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2 years ago

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The number 12 is relatively prime to:_______. a) 6 b) 15 c) 8 d) 25

ra1l [238]

Answer: I think the answer is 15.

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4 years ago

Because of a mistake in packaging, a case of 16 bottles of wine contained 9 of brand A and 7 of brand B, each without labels.

attashe74 [19]

Answer:

a) The probability that all five are brand A is 0.0288

b) The probability that exactly two bottles are brand A is 0.0288

c) The probability that none of the bottles is brand A is 0.0048

Step-by-step explanation:

We have 9 bottles of brand A and 7 bottles of brand B.

The total of bottles is 16.

a) The probability that all five bottles are brand A is given by:

$P(5A)=\frac{9}{16} \frac{8}{15}\frac{7}{14} \frac{6}{13} \frac{5}{12}=\frac{3}{104}=0.0288$

b) Since we have 9 bottles of brand A we calculate the probability of picking two brand A bottles and the we calculate the probability of picking 3 brand B bottles:

$P(2A3B)=\frac{9}{16} \frac{8}{15}\frac{7}{14} \frac{6}{13} \frac{5}{12}=\frac{3}{104}=0.0288$

c) The probability that none of the bottles is brand A is the same as picking 5 brand B bottles:

$P(5B)=\frac{7}{16} \frac{6}{15}\frac{5}{14} \frac{4}{13} \frac{3}{12}=\frac{1}{208}=0.0048$

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3 years ago

Factor j²-13j+42 The Factored Expression =

Ymorist [56]

Answer:

(j - 7)(j - 6)

Step-by-step explanation:

${j}^{2} - 13j + 42 \\ = {j}^{2} - 7j - 6j + 42 \\ = j(j - 7) - 6(j - 7) \\ = (j - 7)(j - 6) \\ are \: required \: factors$

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3 years ago

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