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3 years ago

Heba ate 1/12 of a box of cereal. Now the box is 3/4 full

Mathematics

2 answers:

vaieri [72.5K]3 years ago

7 0

Answer:

5/6

Step-by-step explanation:

1/12 + 3/4 = 9/12

9/12 + 1/12 = 10/12

simplify 10/12

= 5/6

Alona [7]3 years ago

3 0

Heres the answer
Hope it helps

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Please help due tomorrow

mel-nik [20]

Answer: x= 2.5, y = 10

Step-by-step explanation:

<u><em>I'm going to assume that these photocopies are proportional in relations to each other.</em></u>

If they're proportional, you can set up two proportions:

$1) \frac{x}{5} =\frac{3}{6} \\\\2) \frac{5}{y} =\frac{3}{6}$

And cross-multiply:

$1) 6x = 5*3 \\\\2) 3y = 5*6$

Then solved for x and y:

$1) 6x = 15\\x=\frac{15}{6} =\frac{5}{2} =2.5 \\\\2) 3y = 30\\y=\frac{30}{3} =10$

5 0

3 years ago

A sporting good store is offering 30 percent off of the original price(x) of football cleats. The discount will be reduced by an

mylen [45]

(X times 0.3) -7 is the discount

5 0

3 years ago

Kate has 48 softballs. She wants to divide them evenly among b softball bags. Which expression represents how many softballs she

Makovka662 [10]

Answer:

Step-by-step explanation:

48 softballs

Divided among b bags

Divide 48 by b

48 ÷ b

Division expressions can also be written as fractions

Rewrite

48/b

The correct answer is A, 48/b

Hope this helps :)

8 0

3 years ago

An industry representative claims that 10 percent of all satellite dish owners subscribe to at least one premium movie channel.

Greeley [361]

Answer: 1) 0.6561 2) 0.0037

Step-by-step explanation:

We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

, where p =Probability of getting success in each trial.

As per given , we have

The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10

Sample size : n= 4

Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.

1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = $P(X=0)=^4C_0(0.10)^0(1-0.10)^{4}$

$=(1)(0.90)^4=0.6561$

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.

2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.

= $P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037$

∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.

8 0

3 years ago

Pls help only b,d, and f

Leya [2.2K]

Solving for y right?

2y = 3x + 4 - x
2y = 2x + 4
y = 2x + 4 over 2
y = 2 (x + 2) over 2
y = x + 2

y - 3 = 2x - 6 over 2
y - 3 = 2(x - 3) over 2
y - 3 = x - 3
y = x

x - y - 2 = 2(2x + 1)
-y - 2 = 2(2x + 1) - x
-y = 2(2x + 1) - x + 2
y = -2(2x + 1) + x - 2

8 0

3 years ago