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2 years ago

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A boy band has released two albums. Their first album sold 5.9×10^5 copies. Their second album sold 1.3×10^6 copies. How many to

tal copies has this boy band sold?

2 answers:

PIT_PIT [208]2 years ago

4 0

Answer:

The boy band sold 1.89 ×10^6 copies

Step-by-step explanation:

To get the number of copies we must add 5.9×10^5 copies with 1.3×10^6 copies. We can´t add them like that.

To make the addition we have to adjust the power of 10, so they have the same index

In this case we can change 5.9×10^5 to 0.59×10^6 ( we move the comma to the left , and we add 1 power to the scientific quotation.

So, 0.59×10^6

<u />

<u>+ 1.3×10^6 </u>

1.89 ×10^6

prisoha [69]2 years ago

3 0

The band has sold 1,890,000 copies or 1.89 x 10^6

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4. Suppose that Peculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply thr

Viktor [21]

Answer:

Peculiar purples would be more abundant

Step-by-step explanation:

Given that eculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply through a mechanism in which each single bacterial cell splits into four. Time taken for one split is 12 m for I one and 10 minutes for 2nd

The function representing would be

i) $P=P_0 (4)^{t/12}$ for I bacteria where t is no of minutes from start.

ii) $P=P_0 (4)^{t/10}$ for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

a) Here P0 =3, time t = 60 minutes.

i) I bacteria P = $3(4)^{5} =3072$

ii) II bacteria P = $3(4)^{4} =768$

b) Since II is multiplying more we find that I type will be more abundant.

The difference in two hours would be

$3(4)^{10}- 3(4)^{8} =2949120$

c) i) $P=P_0 (4)^{t/12}$ for I bacteria where t is no of minutes from start.

ii) $P=P_0 (4)^{t/10}$ for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

d) At time 36 minutes we have t = 36

Peculiar purples would be

$i) P=3 (4)^{36/12}=192$

The rate may not be constant for a longer time. Hence this may not be accurate.

e) when splits into 2, we get

$P=P_o (2^t)$ where P0 is initial and t = interval of time

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2 years ago

Jane builds a ramp made of a triangular prism and a rectangular prism. What is the volume

Alex787 [66]

Answer:

Correct option: third one -> 11.5 m3

Step-by-step explanation:

To find the volume of the ramp, first we need to find the volume of the rectangular prism and the volume of the triangular prism:

V_rectangular = 4m * 2m * 1m = 8 m3

V_triangular = (2m * 3.5m * 1m) / 2 = 3.5 m3

Now, to find the volume of the ramp, we just need to sum both volumes:

V_total = V_rectangular + V_triangular = 8 + 3.5 = 11.5 m3

Correct option: third one.

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2 years ago

Read 2 more answers

Consider the following regression model: Humidity = β0 + β1Temperature + β2Spring + β3Summer + β4Fall + β5Rain + ε, where the du

FinnZ [79.3K]

Answer:

The regression equation for the winter rainy days is "Humidity = (β0 + β5) + β1Temperature".

Step-by-step explanation:

Given:

Humidity = β0 + β1Temperature + β2Spring + β3Summer + β4Fall + β5Rain + ε ...........(1)

Since there can be only one of spring, summer,fall, and winter at a point in time or in a season, we will have the following when there are winter rainy days:

Spring = 0

Summer = 0

Fall = 0

Rain = 1

Substituting all the relevant values into equation (1) and equating ε also to 0, a reduced form of equation (1) can be obtained as follows:

Humidity = β0 + β1Temperature + (β2 * 0) + (β3 * 0) + (β4 * 0) + (β5 * 1) + 0

Humidity = β0 + β1Temperature + 0 + 0 + 0 + β5 + 0

Humidity = (β0 + β5) + β1Temperature

Therefore, the regression equation for the winter rainy days is "Humidity = (β0 + β5) + β1Temperature".

3 0

2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So 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.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$ .

In which

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0

2 years ago

Adult male heights have a normal probability distribution with a mean of 70 inches and a standard deviation of 4 inches. What is

NNADVOKAT [17]

Answer:

The probability that the man is greater than 74 inches is 0.1587

Step-by-step explanation:

The required probability is found by evaluating the area under the corresponding distribution curve for the corresponding values

The standard normal variate factor (Z) is given by

$Z=\frac{x-\bar {X}}{\sigma }$

where

$\bar{x}$ is mean of the data

$\sigma$ is the standard deviation of the data

Thus corresponding to x = 74 the Z factor equals

$Z=\frac{74-70}{4}=1$

Using the standard normal distribution table corresponding to mean of 70 and deviation of 4 the area under the curve corresponding to Z = 1 equals

0.1587

6 0

2 years ago