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2 years ago

8

Which of the following statements best describes the data?

1 answer:

Svet_ta [14]2 years ago

8 0

Im not for sure but i think its the 1/3 if students joined music

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Mr. Roseanne will buy 12 1/2 pounds beef for for a cookout. The price for the first 3 pounds of beef is $4.99 per pound. After t

Yuri [45]

Answer:

As per the statement:

Mr. Roseanne will buy $12\frac{1}{2}$ pounds beef for for a cookout.

⇒ Weight of the beef he buy = $\frac{25}{2}$ pounds.

It is given that:

The price for the first 3 pounds of beef is $4.99 per pound.

then;

the price for 3 pounds he will pay= $3 \times 4.99 =\$ 14.97$

After that, the price is $4.49 per pound.

Remaining weight of beef left = $\frac{25}{2} -3 = \frac{19}{2}$ pounds.

then;

the price of $\frac{19}{2}$ pounds he will pay = $\frac{19}{2} \times 4.49= \$42.655$

now, total money he will pay for the $12\frac{1}{2}$ pounds beef = $14.97+42.655 = \$57.625$

Therefore, $57.625 will Mr. Roseanne pay for the beef

4 0

3 years ago

If 8, x, 18 form a geometric sequence , find a value for x

katrin [286]

Answer: 13

Step-by-step explanation: It goes up by 5 so 8, 13, 18, 23, and etc

Please tell me if I'm wrong

4 0

2 years ago

Read 2 more answers

Whats .4 (repeating) divided by 3/8??

hichkok12 [17]

.444444 divided by 3/8 = 0.01666666666

6 0

3 years ago

Read 2 more answers

The number of charter schools is 4 less than twice the number of alternative schools and there are 48 charter schools how many a

jarptica [38.1K]

Let x = the number of alternative schools.

Therefore, 2x - 4 = 48.
2x -4 +4 = 48 + 4
2x = 52
x = 26

Therefore, there are 26 alternative schools(You could check this answer by substituting x=26 into the equation above.).

7 0

2 years ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

2 years ago