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2 years ago

7

estimate the sum or difference. 4/5+3/7, round 4/5 to its closest benchmark. round 3/7 to its closest benchmark. add to find the

estimate.

1 answer:

Ann [662]2 years ago

7 0

Answer:

About 1.5

Step-by-step explanation:

We can round 4/5 to 1.

We can round 3/8 to 4/8 = 1/2.

So our estimate is

4/5 + 3/8 =

about 1 + 1/2 = 1.5

(Shh... let's check using the calculator... it's 1.2. We were pretty close!)

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The number of shelves, s, in a bookcase and the number of books, b, the bookcase can hold<br>

Fiesta28 [93]

Answer:

whats the question

Step-by-step explanation:

5 0

2 years ago

30 pts! solve the following equation. (x - 18)^2 = 1

melamori03 [73]

Answer:

D - x = 19 x = -17

Step-by-step explanation:

Factor the equation to make the problem easier - you should have this:

(x - 18) (x + 18) = 1

Set each binomial to 1:

x - 18 = 1

x + 18 = 1

This should give you x = 19 and x = -17

8 0

2 years ago

Read 2 more answers

A student takes a driving test until it is passed.

Leya [2.2K]

Answer:

The probability that the test is taken an even number of times is 0.30.

Step-by-step explanation:

The probability that a student passes the driving test at any attempt is,

<em>p</em> = 4/7.

The event of a student passing in any attempt is independent of each other.

The probability that the test is taken an even number of times is:

P (even number of tests) = P (Passing in the 2nd attempt)

+ P (Passing in the 4th attempt)

+ P (Passing in the 6th attempt) ...

If a student passed in the 2nd attempt it implies that he failed in the first.

Then, P (Passing in the 2nd attempt) = $(\frac{3}{7}) \times (\frac{4}{7})$

Similarly, P (Passing in the 4th attempt) = $(\frac{3}{7})^{3} \times (\frac{4}{7})$ , since he failed in the first 3 attempts.

And so on.

Compute the probability of an even number of tests as follows:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

The result follows a Geometric progression for infinite values.

The sum of infinite GP is:

$S=\frac{a}{1-r^{2}}$

The probability is:

P (even number of tests) = $(\frac{3}{7}) \times (\frac{4}{7})+(\frac{3}{7})^{3} \times (\frac{4}{7})+(\frac{3}{7})^{5} \times (\frac{4}{7})+...$

$=\frac{(\frac{3}{7})(\frac{4}{7} ) }{1-(\frac{3}{7})^{2}}\\=\frac{12}{49}\times\frac{49}{40}\\ =\frac{12}{40}\\ =0.30$

Thus, the probability that the test is taken an even number of times is 0.30.

7 0

3 years ago

⟟ need help with this please someone and please don’t do links

ANEK [815]

B B B B B B B B B B B

8 0

1 year ago

In which of the options below will the number 3 correctly fill in the blank. Select all that apply. A) ___ : 4 = 12 : 16 B) 1 :

aniked [119]

Answer:

A and B

Step-by-step explanation:

Given

List of given options

Required

Which will correctly take 3 to fill the blank

Represent the blanks with x

Option A:

$x : 4 = 12 : 16$

Convert to fractions

$\frac{x}{4} = \frac{12}{16}$

Multiply through by 4

$4 * \frac{x}{4} = 4 * \frac{12}{16}$

$x = \frac{48}{16}$

$x = 3$

Option B:

$1 : 5= x : 15$

Convert to fraction;

$\frac{1}{5} = \frac{x}{15}$

Multiply through by 15

$15 * \frac{1}{5} = \frac{x}{15} * 15$

$15 * \frac{1}{5} = x$

$3 = x$

$x = 3$

Option C:

$x : 1 = 12 : 3$

Convert to fraction

$\frac{x}{1} = \frac{12}{3}$

$x = 4$

Option D

$15:x = 3:1$

Convert to fraction

$\frac{15}{x} = \frac{3}{1}$

Cross Multiply

$3 * x = 15 * 1$

$3 * x = 15$

Divide through by 3

$x = 5$

From the above calculations.

<em>Option A and B can be filled with 3</em>

5 0

2 years ago