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2 years ago

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A student missed 45 problems on a mathematics test and received a grade of 39%. If all the problems were of equal value, how man

y problems were on the test

1 answer:

Virty [35]2 years ago

8 0

There were 74 problems in the test.

<h2>Rounding-off percentage</h2>

Let the student got total number of problems in the test to be X.

Percentage of correct answer = 39 %

⇒ 61% of X were incorrect

⇒61/100 x X = 45

⇒61X = 45x100

⇒X = 4500/61

⇒X=73.77

After rounding off 73.77, we get 74.

Hence, we can say that there were total of 74 problems in the test.

Learn more about percentage rounding-off on:

brainly.com/question/12337260

#SPJ4

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Can someone help me with this? I don't really understand it

SSSSS [86.1K]

Answer:

#1 goes to B.

#2 goes to C.

#3 goes to D.

#4 goes to A.

Hope this helps.

6 0

3 years ago

Suppose f and g are continuous functions such that g(6) = 6 and lim x → 6 [3f(x) + f(x)g(x)] = 45. Find f(6).

aleksandr82 [10.1K]

Since g(6)=6, and both functions are continuous, we have:

$\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5$

if a function is continuous at a point c, then $lim_{x \to c} f(x)=f(c)$ ,

that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.

Thus, since $lim_{x \to 6} f(x)=5$ , f(6) = 5

Answer: 5

7 0

3 years ago

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

Katena32 [7]

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

3 0

3 years ago

(he was flunking school to get out of school plus he worked at kfc and loved his job, during the 11th grade btw)

galben [10]

Use quillbot it can summarize it

4 0

3 years ago

You have decided to purchase a car for $19,000. The credit union requires a 10% down payment and will finance the balance with a

Lady_Fox [76]

<h2>○=> <u>Correct option</u> :</h2>

$\color{plum} \bold{ \tt \: \bold{C. \$20,652.00}}$

<h3>○=> <u>Steps to derive correct option</u> :</h3>

Selling price of a car = $19,000

Percentage of sales tax in the city = 8.3%

Sales tax :

$= \tt \frac{8.3}{10 \times 100} \times 1900$

$= \tt \frac{8.3 \times 19000}{1000}$

$= \tt \frac{1577000}{1000}$

$\color{plum} = \tt \$1577$

Thus, the sales tax on the car = $1577

Cost of license and title = $75

Total price of car :

= Cost price of car + Sales tax + cost of license/title

$= \tt19000 + 1577 + 75$

$\color{plum} = \tt \$20652$

Thus, the total purchase price of the car = $20,652.00

Therefore, the correct option is <em>(C) $20,652.00</em>

6 0

2 years ago