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3 years ago

12

What is the truth value of (~p V q) ^ r if p is true, q is false, and r is true?

1 answer:

disa [49]3 years ago

4 0

Answer:

i dont know im only getting points for this so i dont

fail and my mother hate me more for having to F's and to C's she wants all A's but pls dont get upset at me for not answering this pls it would be helpful

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Find the probability of spinning an even number AND rolling an odd number.

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You have a 1/2 chance of spinning an even number on the spinner, and a 1/2 chance of rolling an odd number on the die, 1/2 times 1/2 give you a 1/4 chance of doing both of those tasks.

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Round to the nearest hundredth. Question is BC=

AveGali [126]

Answer:

BC = 15.01

Step-by-step explanation:

7 x tan(65) = 15.01154844

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Carey creates the table below to help him determine 75% of 36. 75% of 36 25% 25% 25% 25% One-fourth One-fourth One-fourth One-fo

alina1380 [7]

Answer:

The correct option is (A).

Step-by-step explanation:

The expression Carey is solving is:

$X=75\%\ \text{of}\ 36$

Simplify the expression as follows:

$X=75\%\ \text{of}\ 36$

$=\frac{75}{100}\times 36\\\\=\frac{3}{4}\times 36\\\\=3\times 9\\\\=27$

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Thus, the correct option is (A).

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Howard pays $327 for one dozen collector's edition baseball cards. About how much does he pay for each baseball card?

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In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been p

Illusion [34]

Answer:

The right answer is:

(a) 0.1456

(b) 18.125, 69.1202, 8.3139

Step-by-step explanation:

Given:

N = 24

n = 5

r = 7

The improperly drilled gearboxes "X".

then,

⇒ $P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}$

(a)

P (all gearboxes fit properly) = $P(x=0)$

= $\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}$

= $0.1456$

(b)

According to the question,

$X = 91+5$

Mean will be:

⇒ $\mu = E(x)$

$=E(91+5)$

$=9E(1)+5$

$=9.\frac{nr}{N}+5$

$=9.\frac{5.7}{24} +5$

$=18.125$

Variance will be:

⇒ $\sigma^2=Var(X)$

$=V(9Y+5)$

$=81.V(Y)$

$=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}$

$=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}$

$=69.1202$

Standard deviation will be:

⇒ $\sigma = \sqrt{69.1202}$

$=8.3139$

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4 years ago