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2 years ago

Simplify 6r · s · 4rt.

Mathematics

2 answers:

yanalaym [24]2 years ago

5 0

(6r)(s)(4rt)
(6r)(s)(4rt) = 24r²st

Sindrei [870]2 years ago

3 0

The answer would be ~ $24r^2st
$

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Simplify. (4w + 3) – 2w + 7 2w + 10 2w - 10 6w + 10 6w - 10

7nadin3 [17]

Hey there,

2w+10 would be the best choice here.
Sorry if I'm wrong, good luck!

:)

3 0

3 years ago

The price of gasoline at a service station increased from $1.65 per gallon last week to $1.82 per gallon this week. Sally paid $

Mila [183]

Answer:

$2.72

Step-by-step explanation:

The increase from last week is the fraction ...

(new price)/(old price) -1 = (1.82/1.65) -1 = 17/165

so the amount of increase Sally will pay this week is ...

17/165×$26.40 = $2.72

_____

Comment on the working

We could work this several ways. Perhaps most straightforward would be to determine how many gallons Sally bought last week, and multiply that by the change in price. Or, the number of gallons could be multiplied by the new price per gallon and the difference from the old purchase amount found.

If you look at the numbers, you see that we have effectively done it the first way. The change in price is $0.17 per gallon, and the number of gallons Sally bought is $26.40/$1.65.

8 0

2 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) )

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