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3 years ago

Which value is smaller - 0.00032 or 0.00204 - explain using scientific notation

Mathematics

1 answer:

Sedaia [141]3 years ago

7 0

Answer:

The smaller value is 0.00032 ( $(3.2)(10^{-4})$ )

Step-by-step explanation:

1. You have the following values shown in the problem:

0.00032 and 0.00204

2. To write them in scientific notatio you must move the decimal point to the right until you have a non-zero number on the left of the decimal point.

3. Keeping this information on mind, you have:

$0.0032=(3.2)(10^{-4})\\0.00204=(2.04)(10^{-3})$

4. As you can see, the exponent -4 is smaller than the exponent -3, therefore, the smaller value is ( $(3.2)(10^{-4})$ .

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Mr.Chen left $45,000 to his aunt when he died. This was 5% of his total worth. a) how much was Mr.Chen worth when he died? Show

marysya [2.9K]

A) Mr Chen was worth:

45,000/5 = 9,000

9,000 x 100 = 900,000

The answer is 900,00

You can also write this as an equation.

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3 years ago

Find the interval(s) over which the function is increasing

anastassius [24]

The intervals refer to the x values where the function (graph) is increasing which means the y values are getting larger from left to right. Parantheses are used not brackets (parantheses means it doesn't include the value and brackets mean it does) so here the intervals would be
( negative infinity , -3)U( -3, 0 )

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3 years ago

A library has five copies of a certain textbook on reserve of which two copies (1 and 2) are hardcover books and the other three

Gre4nikov [31]

Answer and Step-by-step explanation:

(a) Since the student have to stop when he gets to w softcover book, then the sample space should always stop at either 3 or 4 or 5. If the students examines the books randomly and stops at a softcover book, here are the possible ways he could carry this out,

S = {3, 4, 5, 13, 14, 15, 23, 24, 25, 123, 124, 125, 213, 214, 215}

2. Since only one book is to be examined then the student must check only one softcover book and end the examining. To do this, either he examines just 3 or 4 or 5. Therefore,

A = {3, 4, 5}

3. We need to pick the possibilities from the sample space that shows that the examining ended at book 5. Therefore,

B = {5, 15, 25, 125, 215}

4. We need to pick possibilities from the sample space that shows that book 1 was not examined. Therefore,

C = {3, 4, 5, 23, 24, 25}

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3 years ago

A security fence encloses a rectangular area on one side of a park in a city. three sides of fencing are? used, since the fourth

alexandr1967 [171]

Let x = length of the park
Let y = width of the park

Because the area is 392392 ft², therefore
xy = 392392 (1)
Because three sides of fencing measure 5656 ft, therefore
2x + y = 5656 (2)
That is
y = 5656 - 2x (3)
Substitute (3) into (1).
x(5656 - 2x) = 392392
5656x - 2x² = 392392
2x² -5656x + 392392 = 0
x² - 2828x + 196196 = 0

Solve with the quadratic formula.
x = (1/2)*[2828 +/- √(2828² - 4*196196)]
= 2756.83 or 71.17

Answer:
The possible dimensions are 2756.8 ft and 71.2 ft (nearest tenth)

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3 years ago

Mrs. Teabottle is quite worried about the annual Gardenia Banquet. Ms. Aggravant, Mr. Bellicose, Mrs. Colic and Dr. Dreck all de

nasty-shy [4]

Answer:

Step-by-step explanation:

To abbreviate, I will refer to each person by their initials (A,B,C,D).

We will count all the possibilities depending on C and completing each case.

C can sit on the blue, red or green tables. Suppose that C sits on the blue table. Now, A can sit on the yellow or orange tables, then there are 2 possibilities to sit A. After choosing, there remains only one choice for B, either the yellow or orange table, the one which A didn't sit.

Now, the blue, yellow and orange tables are occupied, so D can sit on the red or green tables. Hence, we have 2 choices for D. In total, we have 2×2=4 ways to sit everyone if we choose blue for C.

Suppose that C sits on the green or red tables (2 choices). Now, we split this in two cases.

If we use the yellow and blue tables for A and B, there are 2 ways of seating them (A to yellow, B to blue, or A to blue, B to yellow). For D, we have 2 choices, the red/green table (depending on C) and the orange table. Thus, in this case, we have 2×2×2=8 ways of seat the guests.

If we don't use precisely the yellow and blue tables for A and B, one of them must sit at the orange table. There are 2 ways of doing this (A in orange or B in orange). The other must sit on the blue or yellow tables (2 choices). Finally, D has only 1 choice, the table not used by C. Then we have 2×2×2=8 ways of doing this.

Therefore, considering all the cases we have 4+8+8=20 ways of seating the guests.

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3 years ago