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3 years ago

How to put one hundred forty one million six hundred twenty thousand in standard form

Mathematics

2 answers:

Brums [2.3K]3 years ago

6 0

Answer:

140,620,000

Step-by-step explanation:

There are two types of standard form. The one above (given to you) is the normal standard form. However, there is another one. This one would look like:

140,620,000 = 1.4062 x 10^8

Remember that in this standard form, you must move the decimal to just after the first place value, and the amount of moves you made for the decimal is the amount in the power over 10.

Ksivusya [100]3 years ago

5 0

Answer: 141,620,000

you have to put each number in its place the 141 in the millions places the 620 in the thousands places and the 0's in the ones tens and hundred places.

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Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

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$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

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$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

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3 years ago

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topjm [15]

Answer:

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Step-by-step explanation:

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Elena L [17]

Answer:

Step-by-step explanation:

Parallel lines have equal slopes, thus

line AB is parallel to line CD

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3 years ago

The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.

nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x). Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.

With reference to table we get

$\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418$

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

$a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85$

and

$b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97$

∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31

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