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3 years ago

9

Sandwich sales for one week are $1,275.00 and 300 sandwiches are sold. How much did each sandwich sell for assuming each sold fo

r the same dollar amount?

2 answers:

ozzi3 years ago

7 0

Answer:

$4.25

Step-by-step explanation:

1275/300=425

Arada [10]3 years ago

4 0

Answer:

$4.75

Step-by-step explanation:

You might be interested in

1. Solve the equation. 2. Check your solution. k÷3=21

Sedaia [141]

Answer:

k = 63

Step-by-step explanation:

3 x 21 = 63

8 0

3 years ago

Which inequality statement best represent the graph?

Vika [28.1K]

The parabola is looking down, so it can be A. or D.
The colored area is inside the parabola, that means that values of y inside the parabola is less than values of y of the parabola.
So,
the answer is D. f(x)< -x²+x-1

4 0

3 years ago

The Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are

almond37 [142]

Answer:

For x = 0, P(x = 0) = 0.35

For x = 1, P(x = 1) = 0.54

For x = 2, P(x = 2) = 0.11

For x = 3, P(x = 3) = 0

Step-by-step explanation:

We are given that the Sky Ranch is a supplier of aircraft parts. Included in stock are 6 altimeters that are correctly calibrated and two that are not. Three altimeters are randomly selected, one at a time, without replacement.

Let X = <u><em>the number that are not correctly calibrated.</em></u>

Number of altimeters that are correctly calibrated = 6

Number of altimeters that are not correctly calibrated = 2

Total number of altimeters = 6 + 2 = 8

(a) For x = 0: means there are 0 altimeters that are not correctly calibrated.

This means that all three selected altimeters are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 3 altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_3$

So, the required probability = $\frac{^{6}C_3}{^{8}C_3}$

= $\frac{20}{56}$ = <u>0.35</u>

(b) For x = 1: means there is 1 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 1 is not correctly calibrated and 2 are correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 2 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_2$

The number of ways of selecting 1 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_1$

So, the required probability = $\frac{^{6}C_2 \times ^{2}C_1 }{^{8}C_3}$

= $\frac{30}{56}$ = <u>0.54</u>

(c) For x = 2: means there is 2 altimeter that is not correctly calibrated.

This means that from three selected altimeters; 2 are not correctly calibrated and 1 is correctly calibrated.

Total number of ways of selecting 3 altimeters from a total of 8 = $^{8}C_3$

The number of ways of selecting 1 correctly calibrated altimeters from a total of 6 altimeters that are correctly calibrated = $^{6}C_1$

The number of ways of selecting 2 not correctly calibrated altimeters from a total of 2 altimeters that are not correctly calibrated = $^{2}C_2$

So, the required probability = $\frac{^{6}C_1 \times ^{2}C_2 }{^{8}C_3}$

= $\frac{6}{56}$ = <u>0.11</u>

(d) For x = 3: means there is 3 altimeter that is not correctly calibrated.

This case is not possible, so this probability is 0.

6 0

3 years ago

In a group of 50 patrons, 12 patrons like lattes and espressos, 9 patrons like espressos and cappuccinos, 8 patrons like lattes

quester [9]

I used a Venn Diagram which I attached.
Think of it as a flower and work your way from the center out to the doubles (two kinds of coffee) and finally the singles (only one kind of coffee)
I place 4 in the center to represent the people that like all three.
Then I put 8 in the Latte Espresso group since they along with the 4 who like all three, make up the 12 who like lattes and espresso. I put 4 in the Latte & Cappuccino group since they and the 4 who like all coffees, make up the 8 who like lattes and cappuccinos. And then I put 5 in the Espresso Cappuccino group who along with the 4 in the middle make up the 9 who like both of those.
In all 20 like lattes and my latte circle already has 16 so I added 4 (who only like lattes). 22 like espresso and I have accounted for 17 (8+4+5) so that means there are 5 who only like espresso. Finally out of the 17 who like cappuccinos, 13 are already accounted for so I will add 4 who like only cappuccinos.
Since there are 50 people and I can account for 34 of them (add all the numbers in all three circles), there must be 50-34 people who don't like any. The correct answer is d.16

3 0

3 years ago

Read 2 more answers

The Rivertown city council is attempting to choose one of four sites (A, B, C, or D) as the location for its new emergency facil

Vikki [24]

Answer:

SITE A

Step-by-step explanation:

Given :

proposed-site Area-Served

1 2 3 4

A 5.2 4.4 3.6 6.5

B 6.0 7.4 3.4 4.0

C 5.8 5.9 5.9 5.8

D 4.3 4.8 6.5 5.1

Area 1 2 3 4

Number-runs 150 65 175 92

Computing the weighted average for the 4 sites :

Site A:

((150*5.2) + (65*4.4) + (175*3.6) + (92*6.5)) / (150 + 65 + 175 + 92)

= 2294 / 482

= 4.7593

Site B:

((150*6.0) + (65*7.4) + (175*3.4) + (92*4.0)) / (150 + 65 + 175 + 92)

= 2344/ 482

= 4.863

Site C:

((150*5.8) + (65*5.9) + (175*5.9) + (92*5.8)) / (150 + 65 + 175 + 92)

= 2819.6/ 482

= 5.850

Site D:

((150*4.3) + (65*4.8) + (175*6.5) + (92*5.1)) / (150 + 65 + 175 + 92)

= 2563.7/ 482

= 5.319

From the weighted average response time computed for the different sites ;

The best location for the emergency facility would be one with the least average response time; which is SITE A.

7 0

3 years ago