Ask question

Ask question

All categories

3 years ago

12

Gary buys a 3 1/2 pound bag of cat food every 3 weeks. Gary feeds his cat the same amount of food each day. What expression coul

d be used to determine the number of pounds of cat food his cat eats each year(1 year-52 weeks)
A. 7/2 x 52/3
B. 7/2 x 3/52
C. 3( 1/2 x 3/52 )
D. 3 ( 1/2 x 52/3 )

2 answers:

Vika [28.1K]3 years ago

8 0

A7/2 x 52/3 i hope that correct

ratelena [41]3 years ago

6 0

Answer: A. $\frac{7}{2}\times\frac{52}{3}$

Step-by-step explanation:

Given: The number of pounds of cat food bought by Gary every 3 weeks. = $3\frac{1}{2}\ pound=\frac{7}{2}\ pound$

If Gary feeds his cat the same amount of food each day.

⇒ The number of pounds of cat food his cat eats each week = $\frac{7}{2\times3}\ pound$

Since 1 year = 52 weeks

Therefore, the number of pounds of cat food his cat eats each year = $52\times\frac{7}{2\times3}=\frac{7}{2}\times\frac{52}{3}$

The expression could be used to determine the number of pounds of cat food his cat eats each year is given by :-

$\frac{7}{2}\times\frac{52}{3}$

You might be interested in

What type of slope is shown in the graph?

boyakko [2]

Answer:

this is a negative slope

Step-by-step explanation:

hope this helps!

6 0

2 years ago

Segment FG begins at point F(-2, 4) and ends at point G(-2, -3). The segment is translated 3 units left and 2 units up, then ref

Alexus [3.1K]

Answer:

$F'G' = 7$

Step-by-step explanation:

Given

$F = (-2,4)$

$G = (-2,-3)$

Required

Distance of F'G'

The transformation that give rise to F'G' from FG are:

Translation
Reflection

The above transformations are referred to as rigid transformation, and as such the side lengths remain unchanged.

i.e.

$F'G' = FG$

Calculating FG, we have:

$FG = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

Where:

$F = (-2,4)$ --- $(x_1,y_1)$

$G = (-2,-3)$ --- $(x_2,y_2)$

$FG = \sqrt{(-2 - -2)^2 + (4 - -3)^2}$

$FG = \sqrt{(0)^2 + (7)^2}$

$FG = \sqrt{0 + 49}$

$FG = \sqrt{49}$

Take positive square root

$FG = 7$

Recall that:

$F'G' = FG$

$F'G' = 7$

6 0

2 years ago

Factor completely.<br> 25 - 64x2

Murljashka [212]

Answer:

Factor 25−64x2

−64x2+25

=(8x+5)(−8x+5)

Answer:

(8x+5)(−8x+5)

3 0

2 years ago

The time a randomly selected individual waits for an elevator in an office building has a uniform distribution with a mean of 0.

Amiraneli [1.4K]

Answer:

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size, of at least 30, can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 0.5, \sigma = 0.289$

What are the mean and standard deviation of the sampling distribution of means for SRS of size 50?

By the Central Limit Theorem

$\mu = 0.5, s = \frac{0.289}{\sqrt{50}} = 0.0409$

The mean of the sampling distribution of means for SRS of size 50 is $\mu = 0.5$ and the standard deviation is $s = 0.0409$

Does it matter that the underlying population distribution is not normal?

By the Central Limit Theorem, since we have of sample of 50, which is larger than 30, it does not matter that the underlying population distribution is not normal.

What is the probability a sample of 50 people will wait longer than 45 seconds for an elevator?

We have to use 45 seconds as minutes, since the mean and the standard deviation are in minutes.

Each minute has 60 seconds.

So 45 seconds is 45/60 = 0.75 min.

This probability is 1 subtracted by the pvalue of Z when X = 0.75. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.75 - 0.5}{0.0409}$

$Z = 6.11$

$Z = 6.11$ has a pvalue of 1

1-1 = 0

0% probability a sample of 50 people will wait longer than 45 seconds for an elevator.

8 0

3 years ago

Merchandise is ordered on November 10; the merchandise is shipped by the seller and the invoice is prepared, dated, and mailed b

natali 33 [55]

Answer:

November 13

Step-by-step explanation:

Following dates are given

On November 10 = Merchandise ordered

Date of an invoice prepared, dated and mailed = November 13

Date when the merchandised received by the buyer = November 18

So, the credit period begins when the invoice is prepared, dated and the mailed by the seller to the buyer as it is the evidence of that the merchandise is ordered

4 0

3 years ago