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Ilia_Sergeevich [38]
3 years ago
6

Evaluate -3+y/5 when y= -2.5

Mathematics
2 answers:
guapka [62]3 years ago
7 0
-1.1. You have to add -2.5 and -3 which gives you -5.5. When you divide -5.5 over 5 that gives you -1.1. 
Thepotemich [5.8K]3 years ago
3 0
(-3-2,5)/5 = (-5,5)/5 = -1,1
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you pick a card at random without getting the first card back you pick a second card at random what is the probability of pickin
Keith_Richards [23]

We have to calculate the probability of picking a 4 and then a 5 without replacement.

We can express this as the product of the probabilities of two events:

• The probability of picking a 4

,

• The probability of picking a 5, given that a 4 has been retired from the deck.

We have one card in the deck out of fouor cards that is a "4".

Then, the probability of picking a "4" will be:

P(4)=\frac{1}{4}

The probability of picking a "5" will be now equal to one card (the number of 5's in the deck) divided by the number of remaining cards (3 cards):

P(5|4)=\frac{1}{3}

We then calculate the probabilities of this two events happening in sequence as:

\begin{gathered} P(4,5)=P(4)\cdot P(5|4) \\ P(4,5)=\frac{1}{4}\cdot\frac{1}{3}=\frac{1}{12} \end{gathered}

Answer: 1/12

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creativ13 [48]

Answer:

x^(2)-3x-16=7x

Step-by-step explanation:

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Read 2 more answers
Rewrite the logarithm as a ratio of common logarithms and natural logarithms. log4 45
tiny-mole [99]

Answer:

Common logarithm

log_451= \frac{log_1_045}{log_1_08}

Natural logarithms

log_445= \frac{log_e45}{log_e4}

or

log_445= \frac{In45}{In4}

Step-by-step explanation:

From the question we are told that

Log_445

Generally converting log from base x to base 10 si mathematically represented as

log_xa=log_1_0a *log_x10

log_xa= \frac{log_1_0a}{log_1_0x}

Therefore

Common logarithm

log_451= \frac{log_1_045}{log_1_08}

Generally Natural logarithms log_ex\ or\ inx is mathematically represented as

log_xa= \frac{log_ea}{log_ex}

Therefore

Natural logarithms

log_445= \frac{log_e45}{log_e4}

or

log_445= \frac{In45}{In4}

6 0
3 years ago
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