Without subtracting 8.5-4.6 for what digit will be in the hundreds place

HELP ME <br> I WILL GIVE YOU BRAINLIEST

lana66690 [7]

76, because the graph shows how many visitors there are around 90 degrees

3 0

2 years ago

The correct scientific notation for the number 500.0 is

marishachu [46]

Answer:

The correct scientific notation for the number 500.0 is $500=5.0\times10^2$

Step-by-step explanation:

While writing a large number in scientific notation we do below mentioned two things.

We take any number between 1 and 10
Then we multiply this number with an exponent of 10.

Hence, any number in scientific notation is in the form $a\times10^b$

Here a is any number between 1 and 10.

b is exponent of 10.

Now, writing 500 in scientific notation, we take a = 5

Hence, we have to take b = 2, since 5 multiplied by 100 equas 500.

Thus, we have

$500=5.0\times10^2$

5 0

3 years ago

According to government data, the probability that a woman between the ages of 25 and 29 was never married is 40%. in a random s

ddd [48]

The probability of finding 2 never married
10c2*0.4^2*0.6^8=
the probability of finding 1 never married
10C1*0.4*0.6^9=
the probability of finding 0 never married
10C0*0.6^10=

the answer is all the above answers added together.
think you can do the rest of the calculation by yourself. :)

btw nCr=n!/(n-r)!r!
good luck

3 0

3 years ago

How many solutions does y=2x+1 and x=-3 have

Nikolay [14]

Answer:

- 5

Step-by-step explanation:

given,

the relation is y=2x+1

if x=-3 than, y =2x+1

=2 * (-3)+1

= -6+1

-5

hence, the required answer is - 5

I found any one solution of given question.

7 0

2 years ago

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

2 years ago