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3 years ago

What's 350/100 as a percent and decimal

Mathematics

2 answers:

yulyashka [42]3 years ago

6 0

Decimal = 3.5
Percent = 350%

To get decimal, do 350

Luden [163]3 years ago

3 0

It would still be 350% and 0.350 because all percents are out of a hundred no matter how high it is.

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Acellus

zhenek [66]

Answer:

y=6 x²+13x-5

Step-by-step explanation:

This is the equation, the constant is -5 instead of 1 ... after calculation

8 0

3 years ago

15x10^2 + 5.2x10^5 can you awnser this

mylen [45]

Answer:

521,500 or 5.215 x 10^5

Step-by-step explanation:

A quick way to solve something that is multiplied by 10^n, simply move the decimal point the same number as n. (For example, 1.6 x 10^3 simply move the decimal to the right 3 times 1600.0)

Therefore,

15 x 10^2 = 1500

5.2 x 10^5 = 520000

1500 + 520000 = 521500 or 5.215 x 10^5

4 0

3 years ago

Read 2 more answers

Please help. Indices and Standard Form question

brilliants [131]

Answer:

A= 4.5×10^11

B= 3.5×10^3

Step-by-step explanation:

(5×10^3)= 5000

(9×10^7)=90000000

multiply both of them

=450000000000 or 4.5×10^11

(7×10^5)=700000

(2×10^2)=200

700000÷200

3500 or 3.5×10^3

8 0

3 years ago

Wriye the mixed number as a fraction 4 1/3

jeyben [28]

A mixed number is a fraction but if you want it as a improper fraction, multiply the denominator by the unit so,
4*3=12
add 1 (from the numerator)
12+1=13

Your fraction would be 13/3 because the denominator stays the same.

3 0

3 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago