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3 years ago

The maximum contribution to an IRA, if you are under the age of 50, is:

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

I just took the quiz !!

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You run around the perimeter of a baseball field at a rate of at most 12 feet per second. Which of the following are possible am

denpristay [2]

Answer: 79 seconds.

Step-by-step explanation:1) we have to calculate the lenght of the baseball field.

perimeter of the baseball field=2(radius)+lenght of a semicircle.

2(radius)=2(175 ft)=350 ft

lenght of a semicircle=40% of a circumference=(40/100)(2πr)

Perimeter of the baseball field=2(175 ft) + (40/100)(2π175 ft)=

=350 ft + 140π ft=350 ft + 439.82 ft=789.82 ft,

2)Now, we calculate the amounts of time that it takes you to ruan arount the baseball field.

10 ft------------------------1 second

789,82 ft----------------- x

x=(789.82 ft * 1 second) / 10 ft=78.98 s≈79 s

7 0

3 years ago

Can you guys plz help me with the customary units of weight

liraira [26]

Are u in 5th grade? Im learning the same thing LOL

6 0

3 years ago

Regrouping add 64 and 9 to show

In-s [12.5K]

73 I hope i helped with the *quashins

5 0

3 years ago

Read 2 more answers

A land surveyor can survey 1 3/5 miles of a land’s borders in 1 ¼ hours. How long will it take him to complete a survey for 1 mi

notsponge [240]

Takes the s 1 h 15 m for 1 3/5 miles
so
75 min divided by 1.6= about 46 min
i am not completely sure if its correct

8 0

3 years ago

A random variable with geometric distribution is a discrete variable that can attain values of 1, 2, . . . . It is used to model

gulaghasi [49]

Answer:

a) You do not have enough to decide the coin is fake

b) You should choose n as the least positive integer such that

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$

Step-by-step explanation:

If the coin is fair, the probability of head = probability of tail = 0.5 .

If the event of tossing the coin follows the <em>Bernoulli's distribution (also called binomial distribution)</em>, then the probability of 1 head in four tosses is

$\binom{4}{1}(0.5)(0.5^3)=4(0.5)^4=0.25=25%$

Since this probability is much less than 50%, it is too early to decide that the coin is not fair.

Suppose you perform the experiment n times in the two scenarios. The probability of having 1 success and n-1 failures in scenario 1 would be

$\binom{n}{1}p_1(1-p_1)^{n-1}$

whereas in scenario 2 would be

$\binom{n}{1}p_2(1-p_2)^{n-1}$

Obviously, you would select scenario 1 if

$\binom{n}{1}p_1(1-p_1)^{n-1}>\binom{n}{1}p_2(1-p_2)^{n-1}$

$p_1(1-p_1)^{n-1}> p_2(1-p_2)^{n-1}$

$\frac{p_1}{p_2}>(\frac{1-p_1}{1-p_2})^{n-1}$

Since the function <em>log(x) is increasing</em>, we can take log on both sides to get

$log(\frac{p_1}{p_2})>(n-1)log(\frac{1-p_1}{1-p_2})$

Since

$01-p_2\Rightarrow \frac{1-p_2}{1-p_1}$

and

$log( \frac{1-p_2}{1-p_1})$

Therefore

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$

So, you should choose n as the least positive integer such that

$n>\frac{p_1/p_2}{log(\frac{1-p_1}{1-p_2})}+1$

8 0

3 years ago