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2 years ago

A, B, C, D, and E are collinear points. AE = 34, BD = 16, and AB = BC =CD. What is the length of CE?

Mathematics

2 answers:

musickatia [10]2 years ago

6 0

Answer:

CE= 18

Step-by-step explanation:

A, B, C, D, and E are collinear points

$AE=AB +BC+CD+DE$

BD = 16, and AB = BC =CD

BD=BC+CD, BC =CD

So $BD= BC+BC$

$16=2BC$

BC= 8

$AB = BC =CD$

AB= 8, BC=8, CD= 8, E= 34

$AE=AB +BC+CD+DE$

$34=8+8+8+DE$

$34=24+DE$ , subtract 24 from both sides

DE= 10

$CE= CD+DE$

$CE= 8+10$

CE= 18

Lesechka [4]2 years ago

5 0

18. Drawn out we know that bc and cd are equal. Bd is 16 so bc and cd are both 8.

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The numbers of regular season wins for 10 football teams in a given season are given below. Determine the range, mean, variance,

tensa zangetsu [6.8K]

We have the following data set:

2,7,15,3,12,9,15,8,3,10

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Then,

$\begin{gathered} \text{Range}=15-2 \\ \text{Range}=13 \end{gathered}$

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$\text{Mean}=\frac{\text{Sum of all data points}}{Number\text{ of data po}ints}$ $\text{Mean}=\frac{84}{10}=8.4$

Population variance formula looks like this:

$\begin{gathered} \sigma^2=\frac{\sum^{}_{}(x-\mu)^2}{N} \\ \text{where,} \\ \sigma^2=\text{population variance} \\ \sum ^{}_{}=addition\text{ of} \\ x=\text{each value} \\ \mu=population\text{ mean} \\ N=\text{ number of values in the population} \end{gathered}$

Then, substituting:

$\begin{gathered} \sigma^2=\frac{(2-14)^2+(3-14)^2+\cdots+(15-14)^2}{10} \\ \sigma^2=20.44 \end{gathered}$

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1 year ago

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2 years ago

Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec

o-na [289]

Answer:

(a) The probability of the event (X > 84) is 0.007.

(b) The probability of the event (X < 64) is 0.483.

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 64.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...$

(a)

Compute the probability of the event (X > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

$=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007$

Thus, the probability of the event (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

$=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483$