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3 years ago

HELP

Mathematics

2 answers:

kicyunya [14]3 years ago

4 0

Answer:

tan

Step-by-step explanation:

take 19 degree as reference angle

using tan rule

tan 19=opposite/adjacent

0.34=14/x

x=14/0.34

x=41.17

x=41.2

gtnhenbr [62]3 years ago

4 0

The answer is C. Tan

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Estimate using the clustering method. 36.61 + 35.907 + 37.2

Elza [17]

Answer:

108

Step-by-step explanation:

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3 years ago

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Given klmn is a trapezoid, kl = mn, m<1 = m<2, lm/kn = 8/9, pklmn = 132. Find: the length of the legs

Ilia_Sergeevich [38]

Answer:

The length of legs lm, mn, and lk is 32 while that of kn is 36.

Step-by-step explanation:

The diagram is attached in the answer.

As both the angles are equal thus the two forms isosceles triangles. This is true because the converse of base angle theorem is applicable here. The theorem clearly states that when a trapezoid is divided by the diagonal, it forms two set of angles where m<1=m<2 and m<lmk=m<nmk

This leads the two set of legs such as

$\dfrac{lm}{kn}=\dfrac{8}{9}$

Or this could be written as

$lm=8x\\kn=9x$

Now the remaining two sides are given as

$lk=lm$ as this form an isosceles triangle so

$lk=8x$

Similarly

$mn=lm\\$

$mn=8x$

Now the perimeter is given as

$\text{perimeter}=lm+mn+kn+lk\\$

Here perimeter is given as 132 so this gives:

$\text{perimeter}=lm+mn+kn+lk\\132=8x+8x+8x+9x\\132=33x\\x=\dfrac{132}{33}\\x=4$

So the four legs are of length as given below:

$lm=8x=8\times 4=32\\mn=8x=8\times 4=32\\lk=8x=8\times 4=32\\kn=9x=9\times4=36$

So the length of legs lm, mn, and lk is 32 while that of kn is 36.

6 0

3 years ago

A die is rolled. If you roll a 1, 2 or 3, you will toss 10 coins. If you roll a 4, 5 or 6, you will toss 20 coins. Let X denote

trasher [3.6K]

Answer:

a) The support of X is {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

b) The mean of X is 7.5

c) The variance of X is 10

Step-by-step explanation:

a) Since you can toss up to 20 coins, and from that you can obtain any number of heads from 0 to 20, then the support of X {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}.

b) To compute the mean of X, we need to see how is the <em>behavior</em> of X. Since 3 dices will gives 10 coins to toss and the other 3 will give us 20, then there is a probability of 1/2 that we tossed 10 coins and a probability of 1/2 that we tossed 20.

The random variable X, conditioned to the event 'The dice is 1,2 or 3' (or equivalently, 10 coins are tossed), will have a binomial distribution with paramenters n = 10, p = 1/2. The mean of X in this case is np = 5. If X is conditioned to the event 'The dice is 4,5 or 6', then X will have also binomial distribution, but this time with paramenters n = 20, p = 1/2. The mean of X in this case is 20*1/2 = 10.

Since each event we conditioned in had probability 1/2 to occur, then E(X) = 1/2 * 5 + 1/2 * 10 = 7.5.

c) Remember that V(X) = E(X²)- E(X)². Since we alredy know the mean of X, we just need to compute the mean of X squared.

The variance of a binomial distribution Z with paramenters n and p is

V(Z) = np(1-p)

since V(Z) = E(Z²)- E(Z)² = E(Z²)- n²p², then

E(Z²) = V(Z) + E(Z)² = np(1-p) + n²p² = p²(n²-n) + np

Therefore

E(X²) = E(X² | 10 coins are tossed) * 1/2 + E(X² | 20 coins are tossed) * 1/2 =

1/2*(0.5²(10²-10) + 5) + 1/2*(0.5²(20²-20) + 10) = 13.75 + 52.5 = 66.25

As a consecuence

V(X) = 66.25 - 7.5² = 10

The variance of X is 10.

5 0

3 years ago

I really really really need help!!!!

Yakvenalex [24]

For $f$ to be continuous at $x=1$ , you need to have the limit from either side as $x\to1$ to be the same.

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(|x-1|+2)=2$
$\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(ax^2+bx)=a+b$

If $a=2$ and $b=3$ , then the limit from the right would be $2+3=5\neq2$ , so the answer to part (1) is no, the function would not be continuous under those conditions.

This basically answers part (2). For the function to be continuous, you need to satisfy the relation $a+b=2$ .

Part (c) is done similarly to part (1). This time, you need to limits from either side as $x\to2$ to match. You have

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(ax^2+bx)=4a+2b$
$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(5x-10)=0$

So, $a$ and $b$ have to satisfy the relation $4a+2b=0$ , or $2a+b=0$ .

Part (4) is done by solving the system of equations above for $a$ and $b$ . I'll leave that to you, as well as part (5) since that's just drawing your findings.

8 0

3 years ago

Write the numerical expression in words two different ways. 7/ (10-3)

pentagon [3]

1) Seven divided by the difference of 10 and 3

2) A fraction where the numerator is seven, and the denominator is the difference of 10 and 3

3) Seven divided by the quantity of 3 subtracted from 10

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3 years ago