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3 years ago

Hello i need some help!

Mathematics

2 answers:

loris [4]3 years ago

8 0

3(2-b) can be rewritten as 3•2 - 3b

In the distributive property you are distributing the number outside the parentheses to the numbers inside.

murzikaleks [220]3 years ago

5 0

Answer:

3×2 + 3×-b

Step-by-step explanation:

$3(2-b)\\= 3\times 2 + 3\times (-b)\\\\= 6 -3b$

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You are at a European beach with 60 other visitors. 36 of them speak English. If you randomly meet two people on the beach, what

Kaylis [27]

Answer:

Assuming I'm one of the 36 English speakers and the other 24 speak Spanish for illustration purposes. The problem can be modeled as 59 marbles with 35 E and 24 S marbles as N = 59 possible outcomes = n(E) + n(S) = 35 + 24.

So I reach into the pile of marbles (on the beach) and the probability that it's p(E) = n(E)/N = 35/59 = 0.593220339 when I meet the one person. ANS

I assume I remember that first person; so I remove him from the marbles (by avoiding him on the beach) and now my probability is p(E and E) = n(E)/N * n(E)-1/(N - 1) = 35/59*34/58 = 0.347749854 ANS

Following the same logic p(E and E and E) = 35/59*34/58*33/57 = 0.201328863 ANS

This last one is different from the first three. This one is p(E >= 1|4 attempts). We can trace out a probability tree to identify those branches that contain at least one E event. So:

EEEE p() = 35/59 * 34/58 * 33/57 * 32/56 =

EEES p() = 35/59 * 34/58 * 33/57 * 24/56 =

EESE p() = 35/59 * 34/58 * 24/57 * 33/56 =

ESEE p() = 35/59 * 24/58 * 34/57 * 33/56 =

SEEE p() = 24/59 * 35/58 * 34/57 * 33/56 =

EESS p() = 35/59 * 34/58 * 24/57 * 23/56 =

ESES p() = 35/59 * 24/58 * 34/57 * 23/56 =

SEES

SESE

SSEE

ESSS And so on, but...a big BUT...why do all this when

SESS

SSES

SSSE

SSSS

p(E>=1|4) = 1 - p(S and S and S and S) = 1 - 24/59 * 23/58 * 22/57 * 21/56 = 0.976652619 ANS. In other words we find the probability of not meeting an Englishman and take 1 minus that value to find the probability of meeting at least one.

Step-by-step explanation:

3 0

2 years ago

Please help it's for my sister

tatiyna

A and d is the one I would pick

3 0

3 years ago

The number of employees at a company

AURORKA [14]

Answer:

This company loses about 12 employees a year

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

A tire manufacturer warranties its tires to last at least 20,000 miles orâ "you get a new set ofâ tires." In itsâ experience, a

Y_Kistochka [10]

Answer:

Probability that a set of tires wears out before 20,000 miles is 0.1151.

Step-by-step explanation:

We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N( $\mu=26,000,\sigma^{2}=5,000^{2}$ )

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = average lasting of tires = 26,000 miles

$\sigma$ = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

P(X < 20,000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20,000-26,000}{5,000}$ ) = P(Z < -1.2) = 1 - P(Z $\leq$ 1.2)

= 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

4 0

3 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago