Ask question

Ask question

All categories

3 years ago

5

I bought 8 shirts that cost an average of $27. My sister then added 3 shirts. The 11 shirts had a mean cost of 28.50. How much d

id my sister’s shirts cost? Need Help by tomarrow can u also put how u got it so I get it

1 answer:

Montano1993 [528]3 years ago

8 0

It cost 50 cent each first i subtracted 27 from 28.50 and it gave me 1.50 de i divided 1.50 by 3 and it gave 50 cent

You might be interested in

I need help on this one asap pliz

Doss [256]

$C(5,4)(.40)^4(1-.40)^1=5(.4)^4(.6)^1=0.0768$

That's 7.68% = 7.7% (rounded)

Note: C(5, 4) is the number of combinations of 5 things taken 4 at a time. You may have seen this written as $_5C_4$

In general, $C(n,r)=\frac{n!}{r!(n-r)!}$

8 0

2 years ago

At which value will the graph of y=tanx have a zero? Please explain the correct answer!

Rina8888 [55]

Answer: None of the above are correct.

Step-by-step explanation:

Given function: $y=\tan x$

We know that the value of tan x =0 at x=0.

Also, $At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{3}=\sqrt{3}$

$At\ x=\frac{\pi}{2},\ \tan\frac{\pi}{2}=\infty$

$At\ x=\frac{\pi}{3},\ \tan\frac{\pi}{4}=1$

Hence, the only option is correct is "None of the above are correct".

4 0

3 years ago

Lauren says the two expressions 3x+4(2x+5) and 6x +5x+5 are equivalent

Allisa [31]

Answer:

so, lauren isn't correct

Step-by-step explanation:

3x + 4 (2x + 5) = 6x + 5x + 5

3x + 8x + 20 = 11x + 5

11x + 20 ≠ 11x + 5

8 0

1 year ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago

Find the Volume of the prism

Strike441 [17]

Answer:

Step-by-step explanation:

volume=area *lenght

for area=2area of triangle +area of rectangle

a=2root(s(s-a)(s-b)(s-c))+7*4 {s is (a+b+c)/2 so s=(3+4+5)/2 finding 5 by pythagoras theorem so s=6}

a=2root(6(6-2)(6-4)(6-5))+28

a=2root48+28=41.85inch^2

now volume=41.86*5 given in fig height is 5

so v=209 inch^3

3 0

2 years ago