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3 years ago

How much is 2 1/2 x 3 3/5

Mathematics

2 answers:

Len [333]3 years ago

6 0

The answer is 8.33333333 I believe

frozen [14]3 years ago

6 0

5/2 * 18/5=90/10=9

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*please help word problem* Adam and Barb had a used book sale over the weekend. Adam sold 4 hardcover books and ten paperbacks a

vekshin1

Let us assume charge for a hardcover book = $ h

And charge for a paperback = $ p.

Adam sold 4 hardcover books and 10 paperbacks and earned $64.

So, we can setup an equation by using above statement.

4h + 10p = 64 -------------equation(1)

Barb sold 5 hardcover books and 7 paperbacks and made $58.

So, we can setup another equation by using above statement.

5h + 7p = 58 -------------equation(2)

Let us solve equation(1) and eqation (2) by elimination method.

Multiplying first equation by -5 and second equation by 4 and adding, we get

-20h -50p = -320

20h +28p = 232

___________________

-22p = -88

Dividing both sides by -22, we get

-22p/-22 = -88/-22

p=4.

Plugging p=4 in first equation.

4h + 10(4) = 64

4h +40 =64

Subtracting 40 from both sides, we get

4h = 24.

Dividing both sides by 4, we get

4h/4 = 24/4

h=6.

Therefore, they charge $6 for a hardcover and $4 for a paperback.

3 0

3 years ago

Ellen read a book a day for 10 weeks. How many books did<br> she read in all?

a_sh-v [17]

1 week = 7 days

10 weeks = 10×7 days

= 70 days

if Ellen reads 1 book in a day

Therefore,

1 day = 1 book

70 days = 70×1

= 70 books.

Thus Ellen reads 70 books in 10 weeks.

Hope helps

8 0

1 year ago

Read 2 more answers

In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo

larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine $z_{\frac{\alpha}{2}}=z_{0.005}$

Now by using z score table we find that $z_{\frac{\alpha}{2}}=2.58$

The boundaries of the confidence interval are:

$\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64$

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

3 0

3 years ago

The school will not announce a snow day unless there is at least 10 inches of snow on the ground. If it has already snowed 1.5 i

matrenka [14]

It needs to snow 8.5 more inches of snow

6 0

3 years ago

Read 2 more answers

How do you find the first five terms of a sequence defined recursively

Blizzard [7]

Interesting question. Good to know for computer science.
Suppose you have a function like
an = 3x - 2 Try the first couple
a1 = 3(1) - 2
a1 = 3 - 2
a1 = 1

a2 = 3(2) - 2
a2 = 6 - 2
a2 = 4 So each term differs by 3
a2 - a1 = 3

an = a_(n - 1) + 3
a3 = a2 + 3
a3 = 4 + 3
a3 = 7

a4 = a3 + 3
a4 = 7 + 3
a4 = 10

a5 = a4+ 3
a5 = 10 + 3
a5 = 13

I'll do one more and then check it.
a6 = a5 + 3
a6 = 13 + 3
a6 = 16

a6 = 3x -2
a6 = 3*6 - 2
a6 = 18 - 2
a6 = 16 which checks.

So the general formula is
an = a_(n - 1) * k if you were multiplying or
an = a_(n - 1) + k if you were adding. The key thing is that you are working with the previous term.

6 0

3 years ago