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iogann1982 [59]
3 years ago
15

Help me please with Math

Mathematics
1 answer:
aksik [14]3 years ago
3 0

Answer:

Question 11 A and D

Step-by-step explanation:

And Question 12 yes They did if you add up history and science together (10) and poetry is 9 and 10>9

Hope this helps!

~R.C aka dj

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What is the value of the expression below when y 8 and z = 8? 9y + 3z​
aleksley [76]

Answer:

96

Step-by-step explanation:

8y+3z

9(8)+3(8)

72+24

96

6 0
3 years ago
A research team at Cornell University conducted a study showing that approximately 10% of all businessmen who wear ties wear the
LenKa [72]

Answer:

a) The probability that at least 5 ties are too tight is P=0.0432.

b) The probability that at most 12 ties are too tight is P=1.

Step-by-step explanation:

In this problem, we could represent the proabilities of this events with the Binomial distirbution, with parameter p=0.1 and sample size n=20.

a) We can express the probability that at least 5 ties are too tight as:

P(x\geq5)=1-\sum\limits^4_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\geq5)=1-(0.1216+0.2702+0.2852+0.1901+0.0898)\\\\P(x\geq5)=1-0.9568=0.0432

The probability that at least 5 ties are too tight is P=0.0432.

a) We can express the probability that at most 12 ties are too tight as:

P(x\leq 12)=\sum\limits^{12}_{k=0} {\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}}\\\\P(x\leq 12)=0.1216+0.2702+0.2852+0.1901+0.0898+0.0319+0.0089+0.0020+0.0004+0.0001+0.0000+0.0000+0.0000\\\\P(x\leq 12)=1

The probability that at most 12 ties are too tight is P=1.

5 0
3 years ago
I’ll be very grateful for anyone’s help I added a photo of the question
Irina18 [472]

Answer:

c

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Hello can someone help me simplify 4) and 5)? Thank you and please include steps :)
Leni [432]
Alrighty

remember some rules
(ab)^c=(a^c)(b^c)
and
x^{-m}=\frac{1}{x^m}
and
x^0=1 for all real values of x
and
(a^b)^c=a^{bc}
and
(\frac{a}{b})^c=\frac{a^c}{b^c}
and
(a/b)/(c/d)=(ad)/(bc)
and
(a^b)(a^c)=a^{b+c}
and
\frac{a^m}{a^n}=a^{m-n}
and don't forget pemdas
example: -x^m=-1(x^m) but (-x)^m=(-1)^m(x^m)

so

4.
(\frac{c^{-2}}{2})^{-2}=

\frac{(c^{-2})^{-2}}{2^{-2}}=

\frac{c^4}{\frac{1}{2^2}}=

\frac{c^4}{\frac{1}{4}}=

4c^4



5.

\frac{(-a)^4bc^5}{-a^2b^{-3}c^0}=

\frac{(-1)^4(a)^4bc^5}{-1(a^2)(\frac{1}{b^3}(1)}=

\frac{(1)a^4bc^5}{\frac{(-1)(a^2)}{b^3}}=

\frac{(a^4bc^5)b^3}{(-1)(a^2)}=

\frac{-a^4b^4c^5}{a^2}=

-a^2b^4c^5
3 0
3 years ago
Jackson is playing a game of chance where he must randomly draw a marble out of a jar. He calculates the probability of drawing
madam [21]

Answer:

12/15

Step-by-step explanation:

If the probability of drawing a read marble is 3/15

Then that would just make the probability of NOT drawing a red marble 12/5

5 0
3 years ago
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