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OverLord2011 [107]

3 years ago

12

Why 1 hundred 4 tens and 14 tens name the same amount.

2 answers:

jeka57 [31]3 years ago

7 0

<span>Because 1 hundred contains 10 tens. So if you are saying 1 hundred and 4 tens it is the same as you would say 10 tens and 4 tens. And if you add them together 10 tens plus 4 tens it will be 14 tens together.</span>

Over [174]3 years ago

5 0

First you need to translate these words mathematically as follows:

1 hundred = 100,

1 ten = 10,

2 tens = 2*10 = 20,

3 tens = 3 * 10 = 30,

4 tens = 4 * 10 = 40,

.

.

10 tens = 10 * 10 = 100

We can express 1 hundred 4 tens and 14 tens as follows:

1 hundred 4 tens = 1 hundred + 4 tens = 100 + 40 = 140 --- (1)

14 tens = 14 * 10 = 140 --- (2)

As (1) = (2), therefore, we can say that 1 hundred 4 tens and 14 tens yield the same amount (which is 140).

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According to an article, 47% of adults have experienced a breakup at least once during the last 10 years. Of 9 randomly select

In-s [12.5K]

Answer:

a) $P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b) $P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

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

c) $P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=9, p=0.47)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Assuming the following questions:

a. exactly five

For this case we can use the probability mass function and we got:

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

b. at least one

For this case we want this probability:

$P(X \geq 1)$

And we can use the complement rule and we got:

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

$P(X=0)=(9C0)(0.47)^0 (1-0.47)^{9-0}=0.0033$

And replacing we got:

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

c. between four and six, inclusive.

For this case we want this probability:

$P(4 \leq X \leq 6)$

$P(X=4)=(9C4)(0.47)^4 (1-0.47)^{9-4}=0.257$

$P(X=5)=(9C5)(0.47)^5 (1-0.47)^{9-5}=0.228$

$P(X=6)=(9C6)(0.47)^6 (1-0.47)^{9-6}=0.135$

And adding we got:

$P(4 \leq X \leq 6)= 0.257+0.228+0.135=0.620$

6 0

3 years ago

12% of 72 is what number?

xxTIMURxx [149]

I hope this helps you ?.12%=72 ?.12=72.100 ?=6.100 ?=600

5 0

3 years ago

Given sin theta = 1/3 and 0 theta pi/2 find tan 2 theta

Kazeer [188]

Answer:

$\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

Step-by-step explanation:

Given: $\sin\theta = \dfrac{1}{3}$

$0\leq \theta \leq \dfrac{\pi}{2}$

Theta lie in first quadrant.

Multiply both sides by 2

$0\leq 2\theta \leq \pi$

2theta lie in I and II quadrant.

$\tan2\theta$ is positive in I and negative in II quadrant.

$\sin\theta=\dfrac{1}{3}$

$\cos\theta=\dfrac{2\sqrt{2}}{3}$

$\sin2\theta=2\sin\theta\cos\theta$

$\sin2\theta=2\cdot \dfrac{1}{3}\cdot \dfrac{2\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{9}$

$\cos2\theta=\pm \sqrt{1-\sin^22\theta}=\pm\dfrac{7}{9}$

$\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}$

$\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

Hence, The value of $\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}$

5 0

3 years ago

Read 2 more answers

darren has 3/4 of a gallon of soil.he needs to plant 5 small plants.if he splits the soil evenly between the plants.how much soi

Natali5045456 [20]

3/4 divides by 5

3/4 x 1/5 = 3/20 of a gallon

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3 years ago

Sarkis wants to draw a triangle with sides measuring 4 ft, 10 ft, and 16 ft. Which is true about Sarkis’s plan?

Liono4ka [1.6K]

Answer:

Sarkis cannot draw a triangle with these side lengths.

Step-by-step explanation:

Sarkis cannot draw a triangle with these side lengths.

5 0

2 years ago