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andreyandreev [35.5K]
3 years ago
8

If there are 460 pounds for every 4 pellets how many pounds are for 16 pellets.

Mathematics
2 answers:
WINSTONCH [101]3 years ago
4 0

Answer:

1840

Step-by-step explanation:

4 pellets x 4 = 16 pellets

460 pounds x 4 = 1840 pounds

HACTEHA [7]3 years ago
3 0

Answer:

1840 pounds for 16 pellets

Step-by-step explanation:

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Micheal buys two bags or chips and three boxes of pretzels for $5.13 . He then buys another bag of chips and 2 boxes of pretzels
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Answer:

So a bag of chip costs $2.39

Step-by-step explanation:

Let x = cost of a bag of chip

Let y = cost of a box of pretzel

Micheal buys two bags of chips and three boxes of pretzels for $5.13

This means

2x + 3y = 5.13 - - - - - - - - - -1

He then buys another bag of chips and 2 boxes of pretzels for $3.09

This means

x + 2y = 3.09 - - - - - - - - - - - -2

Solving equation 1 and equation 2 simultaneously and using the elimination method,

Multiply equation 1 by 1 and equation 2 by 2

2x + 3y = 5.13

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Subtracting,

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5 0
3 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

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Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

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