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IrinaK [193]
3 years ago
7

Plz helppp due today

Mathematics
1 answer:
pshichka [43]3 years ago
3 0

Answer

-2.25 -13/5 1 3/5 -2.25

Step-by-step explanation:

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Assalaamu alaikum...,
aleksandr82 [10.1K]

Answer:

see below

Step-by-step explanation:

alaikum salam.

<u>2. </u><u>find</u><u> </u><u>cos Ф,   sin Ф,  tan Ф</u>

cos Ф = 5 / 13

Ф = 67.38

sin Ф = 12 / 13

Ф = 67.38

tan Ф = 12 / 5

Ф = 67.38

-----------------------

<u>3. find sin β,  cos β,  tan β</u>

sin <u>β</u> = 4 / 5

<u>β</u> = 53.13

cos <u>β</u> = 3 / 5

<u>β</u> = 53.13

tan <u>β</u> = 4 / 3

<u>β</u> = 53.13

4 0
4 years ago
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The moon has a shape of what geometric solid?
Alex777 [14]
<em>The moon is a sphere.</em><span><span><em> </em>
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3 0
4 years ago
7 Mega Middle School had a Valentine
Alex73 [517]

Answer:

600 students attend the school.

Step-by-step explanation:

495-7=488

488/4=122

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8 0
3 years ago
Use the binomial expression (p+q)^n to calculate a binomial distribution with n=5 and p=0.3.(Show all steps)
Helen [10]

Answer:

The binomial in expanded form is (0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}.

Step-by-step explanation:

The Binomial Theorem states that a binomial of the form (a + b)^{n} can be expanded by using the following identity:

(a + b)^{n} = \Sigma \limits^{n}_{k = 0}\,\frac{n!}{k!\cdot (n-k)!}\cdot a^{n-k}\cdot b^{k} (1)

If we know that a = p = 0.3 and n = 5, then the expanded form of the binomial is:

(p+q)^{n} = \frac{243}{100000} + 5\cdot \left(\frac{81}{10000} \right)\cdot q + 10\cdot \left(\frac{27}{1000})\cdot q^{2} + 10\cdot \left(\frac{9}{100} \right)\cdot q^{3} + 5\cdot \left(\frac{3}{10} \right)\cdot q^{4} + q^{5}

(0.3 + q)^{5} = \frac{243}{100000} + \frac{81}{2000}\cdot q + \frac{27}{100}\cdot q^{2} + \frac{9}{10} \cdot q^{3} + \frac{3}{2}\cdot q^{4} + q^{5}

8 0
3 years ago
Which of the following is irrational?<br> 2<br> v1.44<br> 1
kkurt [141]

Answer:

B

Step-by-step explanation:

B

3 0
3 years ago
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