All categories

3 years ago

What is another way to write this number? 300 + 70 + 5 + 8 300 + 70 + 10 100

Mathematics

2 answers:

ololo11 [35]3 years ago

5 0

Answer:

Combine the terms

300 + 70 = 370

5/10 + 8/100

Note that first, you must find common denominators. What you multiply to the denominator, you multiply to the numerator. Multiply 10 to the numerator and denominator of 5/10

(5/10)(10/10) = 50/100

50/100 + 8/100

Combine the terms

50/100 + 8/100 = 58/100

370 58/100 is another way to write it

Yuri [45]3 years ago

3 0

Answer:

Step-by-step explanation:

853 hope this helps thank you so much :)

You might be interested in

Select the correct answer

AlexFokin [52]

<u>Answer:</u>

The correct answer option is C. 56.

<u>Step-by-step explanation:</u>

We are given that a group of 8 friends (5 girls and 3 boys) plans to watch a movie, but they have only 5 tickets.

We are to find the number of combinations these 5 friends could

possibly receive the tickets.

Here, we will use the concept of combination as the order of the friends is not specific.

$5C5+ (5C4 * 3C1) + (5C3*3C2) + (5C2*3C3)$

$=1+5*3 + 10*3 +10*1 = 1+15+30+10=$ 56

5 0

3 years ago

Read 2 more answers

Arithmetic average of real nonnegative numbers is always smaller than or equal to geometric average. Group of answer choices Tru

noname [10]

Answer:

False

Step-by-step explanation:

Actually, the arithmetic average (or mean) is always greater or equal than the geometric average. This is known as the Arithmetic-Geometric inequality (AM inequality). Let a,b be two real numbers, then the AM inequality states that

$\frac{a+b}{2}\geq \sqrt{ab}$

To see that the given statement is false, consider a=1, b=3. The arithmetic mean is equal to (1+3)/2=2, and the geometric mean is equal to $\sqrt{1\cdot 3}=\sqrt{3}$ but $2>\sqrt{3}$ , contrary to the statement (arithmetic>geometric in this case).

4 0

3 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

2 years ago

Which of the following is equivalent to

Firlakuza [10]

Answer: it will be D

Step-by-step explanation: if You graph them you will get the same coordinate plates

7 0

3 years ago

Pls help! ASAP! I will mark Brainliest! If you give me a file or guess I will report you!

madam [21]

Answer:

the second and last one

....................

5 0

3 years ago