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3 years ago

Help I need assistance here

Mathematics

1 answer:

patriot [66]3 years ago

8 0

The ANWSER is unrestricted domain

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(9x5)x10to the 4th power

WARRIOR [948]

$(9\times 5) \times 10^4 = 20 \times 10,000 = 200,000$

So 200,000 is your answer :)

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3 years ago

If the area of a triangle.is 30 square units, what is the height of the triangle if the length of the base of the triangle is 10

Helen [10]

If the area is 30, and the base of the triangle is 10. the way you would find the area is 1/2 base x height. so you would divide 10 by 2 to give you 5. and then ofc 5 x 6 equals 30. but, it’s 10. either way that will give you 30. 10 x 6 = 60, divided by 2 equals 30.

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3 years ago

Which fraction is equivilent to -2/3

Vikki [24]

Answer:

-4/6, -6/9,- 8/12,-10/15

Step-by-step explanation:

hope it helps pls mark brainliest

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3 years ago

Write equation!!<br><br><br> twice the difference of a number and seven equals five

tino4ka555 [31]

Answer:

Its either 2(A+7=5) or 2(A+7) =5

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3 years ago

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

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

5 0

3 years ago