All categories

2 years ago

Suppose fifty students take at least one class from Mr. Green, who teaches band,

Mathematics

1 answer:

Kisachek [45]2 years ago

5 0

Answer:

Step-by-step explanation:

7 students take band and math only

2 students take band and choir only

4 students take choir and math only

We do not include the students that take 3 classes because it says exactly 2 classes

7+2+4 = 13

You might be interested in

erica watch 1/2 of a movie in the morning and she watched some at night and by the end of the day she still had 1/3 of left how

Anastaziya [24]

Convert to equivalent fractions to see
Both bottom s can be 6
So times the half by 3
And times the third by 2
1/2=3/6
1/3=2/6
So she has a total of 5/6 of the movie accounted for...
So the left over is 1/6
1/6 at nighttime

8 0

3 years ago

Point A and point B are placed on a number line. Point A is located at - 20 and

Igoryamba

Answer:

Point B is located on -25

4 0

3 years ago

HELP I NEED HELP WITH THIS

Elena-2011 [213]

Answer:

Step-by-step explanation:

4 0

3 years ago

What is the nth term?

kumpel [21]

Let $a_k$ denote the kth term of the sequence. Then

$a_k=a_1+d(k-1)$

where d is the common difference between consecutive terms in the sequence and a₁ is the first term.

The sum of the first n terms is

$S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n$

From the formula for $a_k$ , we get

$S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)$

$S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k$

$S_n=na_1+\dfrac{d(n-1)n}2$

$S_n=\dfrac n2(2a_1-d+dn)$

So we have $d=-5$ , and $2a_1-d=16$ so that $a_1=\frac{11}2$ .

Then the nth term in the sequence is

$a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}$

7 0

3 years ago

Please round it to the third decimal point.

nasty-shy [4]

Answer:22022635.627

Step-by-step explanation:

photo math

8 0

2 years ago