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3 years ago

35•7 to the 3rd power

Mathematics

1 answer:

dezoksy [38]3 years ago

6 0

Answer:

12005

Step-by-step explanation:

35•7•7•7=12005

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Jody got 90 problems correct on her test. Mildred got 65% of that amount correct. How many did Mildred get correct?

NikAS [45]

Answer:

I think it is 59

Step-by-step explanation:

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3 years ago

A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standa

swat32

All our answers lie in the above statement.

Confidence Level:
The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%.

Margin of Error:
The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .

Confidence Interval:
Lower Limit = 70% - 4% = 66%
Upper Limit = 70% + 4% = 74%

Interpretation:
The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%.

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2 years ago

Ratio of 1/16 to 1/16

aleksklad [387]

Answer 1:16 is the answer

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2 years ago

Read 2 more answers

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

From the above calculations:  $0.267 > 0.03125$  Hence, person B is more likely to get 5

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

Person B has a greater median of 5

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

2 years ago

(c). Under a set of controlled laboratory conditions, the size of the population P of a certain bacteria culture at time t (in s

Bezzdna [24]

(i) Since P(t) gives the population of the culture after t seconds, the population after 1 second is

P(1) = 3•1² + 3e¹ + 10 = 13 + 3e ≈ 21.155

In Mathematica, it's convenient to define a function:

P[t_] := 3t^2 + 3E^t + 10

(E is case-sensitive and must be capitalized. Alternatively, you could use Exp[t]. You can also specify that the argument t must be non-negative by entering a condition via P[t_ ;/ t >= 0], but that's not necessary.)

Then just evaluate P[1], or N[P[1]] or N <at symbol> P[1] or P[1] // N to get a numerical result.

(ii) The average rate of change of P(t) over an interval [a, b} is

(P(b) - P(a))/(b - a)

Then the ARoC between t = 2 and t = 6 is

(P(6) - P(2))/(6 - 2) ≈ 321.030

In M,

(P[6] - P[2])/(6 - 2)

and you can also include N just as before.

(iii) You want the instantaneous rate of change of P when t = 60 (since 1 minute = 60 seconds). Differentiate P :

P'(t) = 6t + 3e^t

Evaluate the derivative at t = 60 :

P'(60) = 6•60 + 3e⁶⁰ = 360 + 3e⁶⁰

The approximate value is quite large, so I'll just leave its exact value.

In M, the quickest way would be P'[60], or you can differentiate and replace (via ReplaceAll or /.) t with 60 as in D[P[t], t] /. t -> 60.

5 0

2 years ago