All categories

3 years ago

Describe two factors in the expression 5(6+4*)

Mathematics

1 answer:

nexus9112 [7]3 years ago

3 0

30x+20 is the simplified form so i don’t know

You might be interested in

Evaluate the expression if a =2.4, b = 0.237, and c =9.49

IRISSAK [1]

Answer:

(0.237)

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Domain:<br> Range:<br> HELP I BEG

dalvyx [7]

Well if the cubic unit is one, then we can assume that the domain is -1<0

7 0

2 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago

Plz Help!! With 5,6,7,8

strojnjashka [21]

6 is 9.088 because it said it for u

8 0

3 years ago

I need help with this uhh points...? <br><br>help

navik [9.2K]

Answer:

Step-by-step explanation:

3 and 1/4 = (3*4 + 1)/4 = 13/4. 3*(13/4) = 39/4 = 9 and 3/4 so it is D

Hope this helps!

5 0

3 years ago