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3 years ago

Divide 33 into 2 groups so the ratio can be 4 to 7

2 answers:

7 0

To do this task the best way is always add both "sides" of ratio and divide whole group by this sum.sum=4+7=1133:11=3Now we know how much is 1 and we can multiply by 4 and by 7 to find the answer.3*4=12 - first group3*7=21 - second group
Your result is ok.

ratelena [41]3 years ago

7 0

4:7....added = 11

4/11(33) = 132/11 = 12
7/11(33) = 231/11 = 21

ur 2 groups are 12:21

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Suppose that the function h is defined, for all real numbers, as follows.

raketka [301]

The function h is defined, for all real numbers , answer is as follows -

h(1) = - 2

h(2) = 3

h(4) = 3.

function, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable

We have been given that

h(x) = 1/4(x) - 1 if x < -2

h(x) = -(x + 1)² + 2 if -2 ≤ x < 2

h(x) = 3 if x ≥ 2

Find h (1), h (2), and h (4)

For h (1) -

Value of h(1) for h(x) = -(x + 1)² + 2 as if -2 ≤ x < 2

So put x = 1

h(x) = -(x + 1)² + 2

h(1) = -(1 + 1)² + 2

h(1) = -4 + 2

h(1) = - 2

For h (2) -

Value of h(2) for h(x) = 3 as if x ≥ 2

h(x) = 3

h(2) = 3

For h (4) -

Value of h(4) for h(x) = 3 as if x ≥ 2

h(x) = 3

h(4) = 3

Hence , the value for given functions - h(1) = - 2 , h(2) = 3 and h(4) = 3.

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5 0

1 year ago

the ratio of potatoes to onions is 3;2 restpectivaly if total number of vegetable is 18 how many potatoes are there

Yuri [45]

It’s 12 potatoes. Hope this helps!

4 0

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

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