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3 years ago

Twenty percent of candies in a package are red. The rest are another color.

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

Among the 20 digits shown, each digit appears in the list twice except 0 and 1 appear 3 times and 6 and 9 appear once. That means ...

1 appears 3 times
2 appears 2 times

So, if 1 and 2 represent red candies, there are 3+2 = 5 red candies in the simulated random sample of 20 candies.

_____

<em>Comment on the question</em>

The simulation makes sense only if it represents taking a single candy from each of 20 packages (of unknown quantity of candies). That is, it seems we cannot answer the question, "how many red candies will be in the packages?" We can only answer the question, "how many of the simulated candies are red?"

timofeeve [1]3 years ago

7 0

Answer:

<em>This is the correct answer! I just took the test! Good luck! </em>

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in 2010 , a city's population was 55,210 and it was declining at a rate of 1.09% each year. which is the best prediction for whe

Anon25 [30]

Answer: 2023

Step-by-step explanation:

Since, the initial population of the city is 55,210,

And, it is declining at a rate of 1.09%.

Thus, the function that shows the population after x years since 2010 is,

$f(x) = 55210(1-\frac{1.09}{100} )^x$

$f(x) = 55210(1-0.0109)^x$

$f(x) = 55210(0.9891)^x$

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⇒ $(0.9891)^x= 0.869407715993$

⇒ $x = 12.769$

Therefore, after 12.769 years the population of town will be equals to 48,000

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⇒ Around 2023 the population will be first below 48,000.

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3 years ago

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Step-by-step explanation:

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2. The Welcher Adult Intelligence Test Scale is composed of a number of subtests. On one subtest, the raw scores have a mean of

IgorC [24]

Answer:

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Step-by-step explaanation:

We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., $\mu$ = 35 and $\sigma$ = 6. We know that the probability density function for a normal distribution with a mean of $\mu$ and a standard deviation of $\sigma$ is given by

$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]$

in this case we have

$f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]$

Let $X$ be the random variable that represents a row score, we find the values we are seeking in the following way

a) we are looking for a number $x_{0}$ such that

$P(X\leq x_{0})$ = $\int\limits^{x_{0}}_{-\infty} {f(x)} \, dx = 0.65$ , this number is $x_{0}$ =37.31

you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)

b) we are looking for a number $x_{1}$ such that

$P(X\leq x_{1})$ = $\int\limits^{x_{1}}_{-\infty} {f(x)} \, dx = 0.9$ , this number is $x_{1}$ =42.70

you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)

c) we find this probability as

$P(28\leq X\leq 38)$ = $\int\limits^{38}_{28} {f(x)} \, dx = 0.57$

you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)

d) we find this probability as

$P(41\leq X\leq 44)$ = $\int\limits^{44}_{41} {f(x)} \, dx = 0.09$

you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)

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