All categories

3 years ago

3. Atile warehouse has Inventory at hand and can put in for a

Mathematics

1 answer:

Nookie1986 [14]3 years ago

5 0

Step-by-step explanation:

Given the equation that represents this order expressed as;

The number of tiles = 12b + 38 where;

b is the the number of bundles ordered

If a customer needs 150 tiles, the total number of bundles ordered can be gotten by simply substituting The number of tiles into the modeled equation and find the value of b. This is as shown below;

On substituting;

150 = 12b + 38

12b = 150 - 38

12b = 112

b = 112/12

b = 9.33

b ≈ 9 bundles

We need to round up the problem because the number of tiles can not be in fraction but as whole numbers.

You might be interested in

Umm, can somebody help me?

Harrizon [31]

Answer:

the answer is 18.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Steven goes to the grocery store and is looking at in Winter squash it has the mass of 1857 grams how many kilograms does the wi

steposvetlana [31]

The answer is 1.857 since there is 1000 grams in a kilogram

6 0

3 years ago

Evaluate the expression 3^4 divided by (14 - 5) x 2<br><br>PLEASE HELP ASAP!

zysi [14]

Answer:

= 18

Step-by-step explanation:

hope this helped

4 0

3 years ago

On Sunday, a local hamburger shop sold a combined total of 540 hamburgers and cheeseburgers. The number of cheeseburgers sold wa

Assoli18 [71]

540 divided by 4 (ratio of hamburgers to cheeseburgers is 3 to 1) = 135. 135 burgers and (135 times 3) 405 cheeseburgers = 540.

6 0

3 years ago

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. If 64 women are

shtirl [24]

Answer:

0.3569 is the probability that they have a mean pregnancy between 266 days and 268 days.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 268 days

Standard Deviation, σ = 15 days

We are given that the distribution of lengths of pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling =

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{64}} = \frac{15}{8}$

P(pregnancy between 266 days and 268 days)

$P(266 \leq x \leq 268) = P(\displaystyle\frac{266 - 268}{\frac{15}{8}} \leq z \leq \displaystyle\frac{268-268}{\frac{15}{8}}) = P(-1.0667 \leq z \leq 0)\\\\= P(z \leq 0) - P(z < -1.067)\\= 0.5000 - 0.1431 = 0.3569 = 35.69\%$

$P(266 \leq x \leq 268) = 35.69\%$

6 0

3 years ago