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4 years ago

Help me again lol im not smart

Mathematics

1 answer:

Artyom0805 [142]4 years ago

6 0

One could be |x|<-2
This could be one since no matter what you put in for x, it will always be positive.

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A study of immunizations among school‑age children in California found that some areas had rates of unvaccinated school‑age chil

Rom4ik [11]

Answer:

Probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

Step-by-step explanation:

We are given that a classroom of 20 children in one such area where 13.5% of children are unvaccinated.

If there are no siblings in the classroom, we are willing to consider the vaccination status of the 2020 unrelated children to be independent.

The above situation can be represented through binomial distribution;

$P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,......$

where, n = number of trials (samples) taken = 20 children

r = number of success = none of the 20 children

p = probability of success which in our case is probability that

children are unvaccinated, i.e; p = 13.5%

Let X = Number of children that are unvaccinated

So, X ~ Binom(n = 20, p = 0.135)

Now, Probability that none of the 20 children in such a classroom would be unvaccinated is given by = P(X = 0)

P(X = 0) = $\binom{20}{0} \times 0.135^{0} \times (1-0.135)^{20-0}$

= $1\times 1 \times 0.865^{20}$

= 0.055

Hence, the probability that none of the 20 children in such a classroom would be unvaccinated is 0.055.

8 0

4 years ago

Find the LCM (least common multiple) of 13 and 52. LCM =

photoshop1234 [79]

52, because 52 is a multiple of 13 (13*4 = 52)

4 0

4 years ago

Read 2 more answers

Suppose that you were not given the sample mean and sample standard deviation and instead you were given a list of data for the

Troyanec [42]

Answer:

So, the sample mean is 31.3.

So, the sample standard deviation is 6.98.

Step-by-step explanation:

We have a list of data for the speeds (in miles per hour) of the 20 vehicles. So, N=20.

We calculate the sample mean :

$\mu=\frac{19 +19 +22 +24 +25 +27 +28+ 37 +35 +30+ 37+ 36+ 39+ 40+ 43+ 30+ 31+ 36+ 33+ 35}{20}\\\\\mu=\frac{626}{20}\\\\\mu=31.3$

So, the sample mean is 31.3.

We use the formula for a sample standard deviation:

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_i-\mu)^2}$

Now, we calculate the sum

$\sum_{i=1}^{20}(x_i-31.3)^2=(19-31.3)^2+(19-31.3)^2+(22-31.3)^2+(24-31.3)^2+(25-31.3)^2+(27-31.3)^2+(28-31.3)^2+(37-31.3)^2+(35-31.3)^2+(30-31.3)^2+(37-31.3)^2+(36-31.3)^2+(39-31.3)^2+(40-31.3)^2+(43-31.3)^2+(30-31.3)^2+(31-31.3)^2+(36-31.3)^2+(33-31.3)^2+(35-31.3)^2\\\\\sum_{i=1}^{20}(x_i-31.3})^2=926.2\\$