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olchik [2.2K]
3 years ago
15

Solve and write in simplest form, 4/6 x 6/8

Mathematics
2 answers:
Aloiza [94]3 years ago
6 0
2/3 x 3/4 because 4 divided by 2 is 2 and 6 divided by 2 is 3 therefore 4/6= 2/3
brilliants [131]3 years ago
5 0

Answer:

1/2

Step-by-step explanation:

4x6=24

6x8=48

24/24=1

48/24=2

Therefore, it is 1/2

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Jada plans to serve milk and healthy cookies for a book club meeting. She is preparing 12 ounces
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Answer:

4n=c

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scoray [572]

A. GIVE ME BRAINLIST

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Read 2 more answers
An someone help please and thx:) !
olganol [36]

Answer:

x = 29

Step-by-step explanation:

All triangle angles add up to 180 so:

x + 17 + 2x + 5 + 3x - 16 = 180

6x + 6 = 180

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3 0
3 years ago
Can each pencil have a mass of 10.5 grams in a pack of 5 pencils weighing 60 to 95 grams?
kvasek [131]
I don't know if I'm right, but if each pencil is 10.5 grams then you multiply it by 5 ( number of pencils) and you get 52.5, which would not meet the standards of a pack weighing 60 to 95 grams.
8 0
3 years ago
In a population of similar households, suppose the weekly supermarket expense for a typical household is normally distributed wi
Rina8888 [55]

Answer:

P(Y ≥ 15) = 0.763

Step-by-step explanation:

Given that:

Mean =135

standard deviation = 12

sample size n  = 50

sample mean \overline x = 140

Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses

Then,

X \sim N ( \mu \sigma)

The probability that X is greater than 140 is :

P(X>140) = 1 - P(X ≤ 140)

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{140-135}{12})

P(X>140) = 1 - P( \dfrac{X-\mu}{\sigma} \leq \dfrac{5}{12})

P(X>140) = 1 - P( Z\leq0.42)

From z tables,

P(X>140) = 1 - 0.6628

P(X>140) = 0.3372

Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140

Then;

Y \sim Binomial (np)

Y \sim Binomial (50,0.3372)

∴

P(Y ≥ 15) = 1- P(Y< 15)

P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )

P(Y \geq 15) = 1 - \begin {pmatrix} ^{50}_0 \end {pmatrix} (0.3372)^0 (1-0,3372)^{50} + \begin {pmatrix} ^{50}_1 \end {pmatrix} (0.3372)^1 (1-0,3372)^{49}  + \begin {pmatrix} ^{50}_2 \end {pmatrix} (0.3372)^2 (1-0,3372)^{48} +...  + \begin {pmatrix} ^{50}_{50{ \end {pmatrix} (0.3372)^{50} (1-0,3372)^{0}

P(Y ≥ 15) = 0.763

7 0
3 years ago
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