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3 years ago

45.15 ÷ 21 pleases solve it for me

Mathematics

1 answer:

sergiy2304 [10]3 years ago

5 0

Answer:

2.15

Step-by-step explanation:

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100 POINTS!!! PLEASE I NEED HELP DUE TODAY!!!!

Volgvan

Using weighed averages, it is found that:

The final grade is 91.
The final grade is 66.8.
The higher grade would be 79.55, with the second grading scheme.
On average, she sold $48,280 per day.
On average, she makes $12.5 per hour.

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To find the weighed average, we multiply each value by it's weight.

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Question 1:

Grade of 91, with a weight of 67%.
Grade of 91, with a weight of 33%.

Thus:

$F = 91\times0.67 + 91\times0.33 = 91$

The final grade is 91.

---------------------------------------

Question 2:

Grade of 83, with a weight of 40%(highest grade).
Grade of 60, with a weight of 30%.
Grade of 52, with a weight of 30%.

Thus:

$F = 83\times0.4 + 60\times0.3 + 52\times0.3 = 66.8$

The final grade is 66.8.

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Question 3:

With teacher 1:

75 with a weight of 25%.
80 with a weight of 10%.
85 with a weight of 40%.
62 with a grade of 25%.

Thus:

$F_1 = 75\times0.25 + 80\times0.1 + 85\times0.4 + 62\times0.25 = 76.25$

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With teacher 2:

75 with a weight of 15%.
80 with a weight of 10%.
85 with a weight of 60%.
62 with a weight of 15%.

Thus:

$F_2 = 75\times0.15 + 80\times0.1 + 85\times0.6 + 62\times0.15 = 79.55$

The higher grade would be 79.55, with the second grading scheme.

---------------------------------------

Question 4:

Average of $36,432, with a weight of $\frac{3}{3+10} = \frac{3}{13}$
Average of $51,834, with a weight of $\frac{10}{13}$

Thus:

$A = 36432\frac{3}{13} + 51834\frac{10}{13} = \frac{36432\times3 + 51834\times10}{13} = 48280$

On average, she sold $48,280 per day.

---------------------------------------

Question 5:

Average of $14.84, with a weight of $\frac{6}{6+8} = \frac{6}{14} = \frac{3}{7}$
Average of $10.76, with a weight of $\frac{4}{7}$

Thus:

$A = 14.84\frac{3}{7} + 10.76\frac{4}{7} = \frac{14.84\times3 + 10.76\times4}{7} = 12.5$

On average, she makes $12.5 per hour.

A similar problem is given at brainly.com/question/24398353

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2

kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed $a\leq x \leq b$ and in your case $\nexists x\in\mathbb{Z}: a\leq x\leq b$ for $a>b$

In fact, considering a set $S$ we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing$ that satisfy $S = S \cup \varnothing$

This means that, by definition $\sum_{x=a}^b \varnothing = 0$

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

we have other problems. Actually, this case is really bad.

Note that $\cos^2(\infty)$ has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$ , the cosine function oscillates between $[-1, 1]$ , and therefore it is undefined. Thus, we cannot evaluate

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0

3 years ago

The scheme where you can find the greatest common divisor of two integers by repetitive application of the division algorithm is

HACTEHA [7]

Answer:

False

Step-by-step explanation:

Euclidean Algorithm is the algorithm that allows us to find the greatest common divisor (gcd) of two integers by repetitive application of the division algorithm.

A division algorithm is an algorithm which, given two integers N and D, computes their quotient and/or remainder.

Quotient and/or Remainder = $\frac{N}{D}$

8 0

3 years ago

Factor the expression 18x^2 - 8

diamong [38]

Answer:

2(3x-2(3x+2)

Step-by-step explanation:

1) 18x^2-8

2) 2(9x^2-4)

3) 2(3x-2)(3x+2)

6 0

3 years ago

A local Honda dealership collects data on customers. Here is data from 311 customers who purchased a Honda Civic. Hybrid Honda C

Kobotan [32]

Answer:

Correct option is: Women are less likely to purchase a Honda Civic with a standard engine than men.

Step-by-step explanation:

Denote the variable as follows:

HHC = Hybrid Honda civic

SHC = Standard-engine Honda Civic

M = male

F = female

The data provided is:

HHC SHC Total

M 77 117 194

F 34 83 117

Total 111 200 311

Compute the probability distribution as follows:

$P (M\cap HHC)=\frac{77}{311}=0.2476$

$P(M\cap SHC)=\frac{117}{311}= 0.3762$

$P(F\cap HHC)=\frac{34}{311}= 0.1093$

$P(F\cap SHC)=\frac{83}{311}= 0.2669$

$P(M) =\frac{194}{311}= 0.6238$

$P(F)=\frac{117}{311} =0.3762$

$P(HHC)=\frac{111}{311}= 0.3569$

$P(SHC)=\frac{200}{311}= 0.6431$

The probability table is:

HHC SHC Total

M 0.2476 0.3762 0.6238

F 0.1093 0.2669 0.3762

Total 0.3569 0.6431 1.0000

Consider the probability distribution above.

The proportion of women purchasing a standard engine Honda civic is less then that for men.

Thus, the correct option is: Women are less likely to purchase a Honda Civic with a standard engine than men.

6 0

3 years ago