A bland dinner would not be <br>a) expensive<br>b) flavorful<br>c) filling<br>d) healthy

a large candy bar was 116 calories. the smaller candy bar was 28% less calories. how many calories does the smaller candy bar ha

Nastasia [14]

Answer: 83.52 calories

Step-by-step explanation:

first candy bar=116 calories

second candy bar= has 28% less calories

to find out the answer we will 28/100=0.28

116x0.28=32.48 calories(less)

116-32.48=83.52 calories for the second candy bar

3 0

3 years ago

What is the great common factor of 24 and 9?

Nezavi [6.7K]

The Greates Common Factor of 24 and 9 is 3

8 0

3 years ago

Read 2 more answers

Hhhhhhhhhhhheeeeeeeeeelllllpppppppppp me

anzhelika [568]

Answer:

m<b=107º

Step-by-step explanation:

180-41-32=107

D and B are congruent

3 0

3 years ago

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in

Marina86 [1]

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

3 0

3 years ago

A firm offers routine physical examinations as part of a health service program for its employees. The exams showed that 8% of t

Alexxandr [17]

Answer: A. 0.20

Step-by-step explanation:

Let A be the event of employees needed corrective shoes and B be the event that they needed major dental work .

We are given that : $P(A)=0.08\ ;\ P(B)=0.15\ ;\ P(A\cap B)=0.03$

We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Then, $P(A\cup B)=0.08+0.15-0.03= 0.20$

Hence, the probability that an employee selected at random will need either corrective shoes or major dental work : $P(A\cup B)= 0.20$

hence, the correct option is (A).

6 0

4 years ago

A bland dinner would not be a) expensiveb) flavorfulc) fillingd) healthy