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Natasha_Volkova [10]
3 years ago
13

1. Burger Town sells cheeseburgers for $7.95 per cheeseburger, c."

Mathematics
1 answer:
labwork [276]3 years ago
6 0

Answer:Idk

Step-by-step explanation:

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Can somebody please help me with the question below and please explain how to find the answer? plz, I'm studying for a test plzz
blagie [28]

Answer:

-10/9 < -2/3 < 4/6 < 6/5 < 5/4

(hope this helps!)

Step-by-step explanation:

3 0
2 years ago
ΔABC is similar to ΔMNO. The scale factor from ΔMNO to ΔABC is 3∕2 . If the area of ΔMNO is 10 square units, what's the area of
Andrews [41]

Answer:

The area of ΔABC= 6.667 square units

Step-by-step explanation:

ΔABC is similar to ΔMNO.

The scale factor from ΔMNO to ΔABC is 3∕2

the area of ΔMNO is 10 square units,

The area of ΔABC/the area of ΔMNO

= 2/3

The area of ΔABC/10= 2/3

The area of ΔABC= 2/3 * 10

The area of ΔABC= 20/3

The area of ΔABC= 6 2/3

The area of ΔABC= 6.667 square units

6 0
3 years ago
ASAP
Virty [35]

Answer:a: y=4.5+0.75x

B: y=2.5+1.25x

C: 7.5

Step-by-step explanation:

6 0
3 years ago
The American Bankers Association reported that, in a sample of 120 consumer purchases in France, 48 were made with cash, compare
tatuchka [14]

Answer:

Step-by-step explanation:

Hello!

You have the information for two variables

X₁: Number of consumer purchases in France that were made with cash, in a sample of 120.

n₁= 120 consumer purchases

x₁= 48 cash purchases

p'₁= 48/120= 0.4

X₂: Number of consumer purchases in the US that were made with cash, in a sample of 55.

n₂= 55 consumer purchases

x₂= 24 cash purchases

p'₂= 24/55= 0.4364

You need to construct a 90% CI for the difference of proportions p₁-p₂

Using the central limit theorem you can approximate the distribution of both sample proportions p'₁ and p'₂ to normal, so the statistic to use to estimate the difference of proportions is an approximate standard normal:

[(p'₁-p'₂) ± Z_{1-\alpha /2} * \sqrt{\frac{p'_1(1-p'_1)}{n_1} +\frac{p'_2(1-p'_2)}{n_2} }]

Z_{0.95}= 1.648

[(0.4-0.4364)±1.648 * \sqrt{\frac{0.4(1-0.4)}{120} +\frac{0.4364(1-0.4364)}{55} }]

[-0.1689;0.0961]

The interval has a negative bond, it is ok, keep in mind that even tough proportions take values between 0 and 1, in this case, the confidence interval estimates the difference between the two proportions. It is valid for one of the bonds or the two bonds of the CI for the difference between population proportions to be negative.

I hope this helps!

8 0
3 years ago
A system of linear equations is graphed below
Mashcka [7]

Answer: D

Step-by-step explanation: 20,20

6 0
2 years ago
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