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1 year ago

If you order a large pizza for $15.29 what is the price per inch of pizza if it is 14 inches?

Mathematics

1 answer:

oksano4ka [1.4K]1 year ago

6 0

The pizza is 1.09$ per inch because is 15.29 divided by 14 I got 1.092 but you can’t round up to 1.10 so it stays 1.09

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What matrix results from −8·A?

djverab [1.8K]

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3 years ago

Which of the functions have a range of all real numbers? Check all that apply. A. y = sec x B. y = csc x C. y = tan x D. y = cot

Nastasia [14]

Answer:

Option C and D that is y= tan x and y= cot x

Step-by-step explanation:

The range is the values of y means the values that y can take in a function:

So, range of tan x and cot x are all real numbers.

Period of tan x is $\pi$

Period of cot x is $\pi$

Option A and B that is secx and csc x are discarded

4 0

4 years ago

Read 2 more answers

A jar contains 12 red marbles numbered 1 to 12 and 6 blue marbles numbered 1 to 6. A marble is drawn at random from the jar. Fin

lora16 [44]

Answer:

$(a)P(Red)=\dfrac{2}{3}\\(b)P(Odd) =\dfrac{1}{2}\\(c)P(\text{Red or Odd Numbered})=\dfrac{5}{6}\\(d)P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

Step-by-step explanation:

Number of Red Marbles{1,2,3,4,5,6,7,8,9,10,11,12},n(R)=12

Number of Blue Marbles{1,2,3,4,5,6},n(B)=6

Total Number of Marbles, n(S)=6+12=18

(a)Probability that the Marble is Red

$P(R)=\dfrac{n(R)}{n(S)} =\dfrac{12}{18} =\dfrac{2}{3}$

(b)Probability that the marble is odd-numbered.

Number of Odd-Numbered Balls, n(O)=9

$P(Odd)=\dfrac{n(O)}{n(S)} =\dfrac{9}{18} =\dfrac{1}{2}$

(c)Probability that the marble is red or odd-numbered.

n(Red)=12

n(Odd Numbered marbles)=9

n(Red and Odd Numbered Marbles)=6

$P(\text{Red or Odd Numbered})=P(Red)+P(Odd\:Numbered)-P(\text{Red and Odd Numbered)}\\=\dfrac{12}{18} +\dfrac{9}{18}-\dfrac{6}{18} =\dfrac{15}{18}\\P(\text{Red or Odd Numbered})=\dfrac{5}{6}$ (d)Probability that the marble is blue or even-numbered.

n(Blue)=6

n(Even Numbered marbles)=9

n(Blue and Even Numbered Marbles)=3

$P(\text{Blue or Even Numbered})=P(Blue)+P(Even\:Numbered)-P(\text{Blue and Even Numbered)}\\=\dfrac{6}{18} +\dfrac{9}{18}-\dfrac{3}{18} =\dfrac{12}{18}\\P(\text{Blue or Even Numbered})=\dfrac{2}{3}$

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eimsori [14]

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