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3 years ago

In triangle KLM, LM =7 and m

Mathematics

2 answers:

hichkok12 [17]3 years ago

5 0

Answer:

KL=7√2

Step-by-step explanation:

LM=7

sin45°=LM/KL

√2/2=7/KL

KL=2*7/√2=14√2/2=7√2

KL=7√2

Radda [10]3 years ago

3 0

Answer:

7sqrt(2)

Step-by-step explanation:

LM squared is 49, and since the measure of angle K is 45, LM is equal to KM. So KM squared is also 49. 49+49=98. The square root of 98 is 7sqrt(2).

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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

vagabundo [1.1K]

Answer:

a) 0.222

b)0.037

c)0.549

Step-by-step explanation:

Let's start defining the random variable ⇒

X : ''The number of missing pulses''

X can be modeled as a Poisson random variable.

X ~ Po(λ)

In a Poisson distribution : μ = λ

Where μ is the mean of the variable.

X ~ Po(0.3)

The probability function for a Poisson random variable is :

In the equation I replace λ = m

$P(X=x)=\frac{e^{-m}m^{x}}{x!}$

Where P(X=x) is the probability of the random variable X to assume the value x

e is the euler number

m = λ is the mean of the variable

In this exercise :

$P(X=x)=\frac{e^{-0.3}0.3^{x}}{x!}$

is the probability function.

For a)

$P(X=1)=\frac{e^{-0.3}0.3^{1}}{1!}=0.222$

For b)

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-[P(X=0)+P(X=1)]$

$P(X\geq 2)=1-(e^{-0.3}+0.222)\\P(X\geq 2)=0.037$

Let's define A :''a disk doesn't contain a missing pulse''

We are looking for P(A1∩A2) of two different disk don't have a missing pulse.

Because of the independence we can write this probability as

P(A1∩A2)= P(A1).P(A2)

The probability of a random disk to don't have a missing pulse is P(X=0)

$P(X=0)=e^{-0.3}$ ⇒

$P(A1).P(A2)=[P(X=0)].[P(X=0)]$

$[P(X=0)].[P(X=0)]=(e^{-0.3})(e^{-0.3})=0.549$

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3 years ago

How many months in 3 million years?

Ilia_Sergeevich [38]

1 year= 12months
3 million years= 3,000,000*12months
=36,000,000 months

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3 years ago

How do you write 26 thousands 13 hundreds on a place value chart?

Sergeeva-Olga [200]

Take your pick of the ways shown in the attachment.

6 0

3 years ago

3/2 x 4/3 x 5/4… x 2006/2005

Lady_Fox [76]

Answer:

1003

Step-by-step explanation:

The problem is a classic example of a telescoping series of products, a series in which each term is represented in a certain form such that the multiplication of most of the terms results in a massive cancelation of subsequent terms within the numerators and denominators of the series.

The simplest form of a telescoping product $a_{k} \ = \ \displaystyle\frac{t_{k}}{t_{k+1}}$ , in which the products of <em>n</em> terms is

$a_{1} \ \times \ a_{2} \ \times \ a_{3} \ \times \ \cdots \times \ a_{n-1} \ \times \ a_{n} \ = \ \displaystyle\frac{t_{1}}{t_{2}} \ \times \ \displaystyle\frac{t_{2}}{t_{3}} \ \times \ \displaystyle\frac{t_{3}}{t_{4}} \ \times \ \cdots \ \times \ \displaystyle\frac{t_{n-1}}{t_{n}} \ \times \ \displaystyle\frac{t_{n}}{t_{n+1}} \\ \\ \-\hspace{5.55cm} = \ \displaystyle\frac{t_{1}}{t_{n+1}}.$ .

In this particular case, $t_{1} \ = \ 2$ , $t_{2} \ = \ 3$ , $t_{3} \ = \ 4$ , ..... , in which each term follows a recursive formula of $t_{n+1} \ = \ t_{n} \ + \ 1$ . Therefore,

$\displaystyle\frac{t_{2}}{t_{1}} \times \displaystyle\frac{t_{3}}{t_{2}} \times \displaystyle\frac{t_{4}}{t_{3}} \times \cdots \times \displaystyle\frac{t_{n}}{t_{n-1}} \times \displaystyle\frac{t_{n+1}}{t_{n}} \ = \ \displaystyle\frac{3}{2} \times \displaystyle\frac{4}{3} \times \displaystyle\frac{5}{4} \times \cdots \times \displaystyle\frac{2005}{2004} \times \displaystyle\frac{2006}{2005} \\ \\ \-\hspace{5.95cm} = \ \displaystyle\frac{2006}{2} \\ \\ \-\hspace{5.95cm} = 1003$

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Marshall's Taffy Shop made 21 kilograms of taffy in 2 days. How much taffy, on average, did the shop make per day?

Mademuasel [1]

Ma yasakō lāgi māphī cāhanchu.

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