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3 years ago

Did I put the decimal in the right place

Mathematics

2 answers:

il63 [147K]3 years ago

5 0

Yes you did, you count up all the decimal spaces in all (which is 4 in this case) and put that amount of decimal spaces in the product.

Katena32 [7]3 years ago

3 0

Yes you did, because you have to move the decimal the number of places added together (How many decimal places?) 0.14 and 0.77 is 4 decimal places, so you move it 4 times to the right at the final product. Hope I helped!

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Of the students at I.S. 59, 230 signed up for the school picnic. That was 50% of all students in the school. How many students a

Nastasia [14]

Answer: 460 students

Step-by-step explanation:

Based on the information given in the question, let the total number of students that attend I.S. 59 be represented by x.

Since we've been given the information that 230 signed up for the school picnic which was 50% of all students in the school. Then, the students in the school will be:

= 50% of x = 230

0.5 × x = 230

0.5x = 230

x = 230/0.5

x = 460

Therefore, 460 students attend the school.

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2 years ago

James is selecting a marble. James chooses a marble at random and then replaces it. He then selects a second marble at random. W

tresset_1 [31]

Answer:

The probability of selecting a solid black marbles both times;

P = 9/100

Attached is the completed question;

Step-by-step explanation:

Number of solid black marbles = 3

Total number of marbles = 10

The probability of selecting a solid black marble;

P1 = 3/10

With the assumption that the marbles are replaced before next selection.

The probability of selecting a solid black marbles both times;

P = P1 × P1 = 3/10 × 3/10 = 9/100

P = 9/100

7 0

3 years ago

A rectangle has an area of 19.38 cm2. When both the length and width of the rectangle are increased by 1.50 cm, the area of the

Lostsunrise [7]

Answer: $5.7\ cm$

Step-by-step explanation:

Given

Rectangle has an area of $19.38\ cm^2$

Suppose rectangle length and width are $l$ and $w$

If each side is increased by $1.50\ cm$

Area becomes $A_2=35.28\ cm^2$

We can write

$\Rightarrow lw=19.38\quad \ldots(i)\\\\\Rightarrow (l+1.5)(w+1.5)=35.28\\\Rightarrow lw+1.5(l+w)+1.5^2=35.28\\\text{use (i) for}\ lw\\\Rightarrow 19.38+1.5(l+w)=35.28-2.25\\\Rightarrow l+w=9.1\quad \ldots(ii)$

Substitute the value of width from (ii) in equation (i)

$\Rightarrow l(9.1-l)=19.38\\\Rightarrow l^2-9.1l+19.38=0\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{(-9.1)^2-4(1)(19.38)}}{2\times 1}\\\\\Rightarrow l=\dfrac{9.1\pm\sqrt{5.29}}{2}\\\\\Rightarrow l=\dfrac{9.1\pm2.3}{2}\\\\\Rightarrow l=3.4,\ 5.7$