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3 years ago

There are 15 questions in this quiz. Im sorry if im putting too much pressure on yall

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

5 0

Answer:

It is 89

Step-by-step explanation: it is really 88.57 but it ask whats it closest to and that would be 89

nika2105 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

i dont really kbow why but i think it 86

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Evaluate the degree of the polynomial (y3 – 2) ( y2 + 11), expand using algebraic expressions and answer.

posledela

Answer:

y^5+11y^3-2y^3-22

Step-by-step explanation:

5 0

3 years ago

If p(x)=x^2+x+1 and q(x)=3x^2-1, find p(7)

Delvig [45]

Answer:

p(7) = 57

Step-by-step explanation:

Because this is a function, all you have to do is input 7 for x like so:

p(x)=x^2+x+1 ---------------> p(7)=7^2+7+1

so 49 + 8

= 57

7 0

2 years ago

Can someone please help me with this?

vovikov84 [41]

Remember that X will always be one up or down depending on if there is a - or not (if it is just an x) and one over! Hope this helps

6 0

3 years ago

Can somebody help me with this?

Ghella [55]

Answer:

x=11

Step-by-step explanation:

Since this is a 90º angle, 46º+(4x)º has to equal 90º, and you can make and solve an equation for this:

46+4x=90

4x=44

x=11

Check:

46+4x=90

46+4(11)=90

46+44=90

90=90

6 0

3 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$

Hence, probability of the stronger team winning overall is:

$=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103$

#Define Y as the random variable for winning 2/3 games.:

$P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480$

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0

3 years ago