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3 years ago

Can someone help me with question 11

Mathematics

1 answer:

Aleksandr-060686 [28]3 years ago

6 0

Answer:

no i can not

Step-by-step explanation:

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-5f - 7g + f - 5g - 4 + f

AfilCa [17]

The answer is -12g-3f-4. Hope this helps!

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3 years ago

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Nate's classroom has three tables of different lengths. One has a length of 4 1/2 feet, another has a length of 4 feet, and a th

Anarel [89]

12 feet.
4 1/2= 4.5
4 1/2+ 4= 9.5 + 2.5 = 12

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3 years ago

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You have two biased coins. Coin A comes up heads with probability 0.1. Coin B comes up heads with probability 0.6.However, you a

Andrews [41]

Answer:

The probability that our guess is correct = 0.857.

Step-by-step explanation:

The given question is based on A Conditional Probability with Biased Coins.

Given data:

P(Head | A) = 0.1

P(Head | B) = 0.6

By using Bayes' theorem:

$P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)}$

We know that P(B) = 0.5 = P(A), because coins A and B are equally likely to be picked.

Now,

P(Head) = P(A) × P(head | A) + P(B) × P(Head | B)

By putting the value, we get

P(Head) = 0.5 × 0.1 + 0.5 × 0.6

P(Head) = 0.35

Now put this value in $P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)}$ , we get

$P(B|Head) = P(Head|B) \times \frac{P(B)}{P(Head)}$

$P(B|Head) = 0.6 \times \frac{0.5}{0.35}$

$P(B|Head) = 0.857$

Similarly.

$P(A|Head) = 0.857$

Hence, the probability that our guess is correct = 0.857.

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3 years ago

There are

Archy [21]

17 books in total, and 8 new. To find the number of used books, we subtract 8 from 17.

17-8= 9 used books

Now we have 8 new books and 9 used books.

All books: 17
Used books: 9
New books: 8

(a) all books: used books
17: 9

(b) new books: used books
8: 9

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3 years ago

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In the right ΔABC, AN is the altitude to the hypotenuse. Find BN, AN, and AC, if AB=2√5 in and NC= 1 in.

kipiarov [429]

Answer:

$BN=4\ \text{in}$

$AN=2\ \text{in}$

$AC=\sqrt{5}\ \text{in}$

Step-by-step explanation:

Let $BN=x$

$BC=c$

$NC=y=1\ \text{in}$

$AB=a=2\sqrt{5}\ \text{in}$

$AC=b$

We have the relation

$\dfrac{BC}{AB}=\dfrac{AB}{BN}\\\Rightarrow \dfrac{c}{a}=\dfrac{a}{x}\\\Rightarrow c=\dfrac{a^2}{x}\\\Rightarrow x+1=\dfrac{a^2}{x}\\\Rightarrow x^2+x=20\\\Rightarrow x^2+x-20=0\\\Rightarrow x=\frac{-1\pm \sqrt{1^2-4\times 1\times \left(-20\right)}}{2\times 1}\\\Rightarrow x=4,-5$

$\boldsymbol{BN=4\ \text{in}}$

$h=\sqrt{a^2-x^2}\\\Rightarrow h=\sqrt{(2\sqrt{5})^2-4^2}\\\Rightarrow h=2\ \text{in}$

$\boldsymbol{AN=2\ \text{in}}$

$b=\sqrt{h^2+y^2}\\\Rightarrow b=\sqrt{2^2+1^2}\\\Rightarrow b=\sqrt{5}\ \text{in}$

$\boldsymbol{AC=\sqrt{5}\ \text{in}}$

8 0

3 years ago