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3 years ago

Solve for d 3(d+12)=8-4d ?????

Mathematics

1 answer:

zysi [14]3 years ago

5 0

Answer:

d=-4

Step-by-step explanation:

the answer is negative 4 have a nice day :)

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the ratio of girls to boys in the sixth grade is 5:3. If there are 400 sixth graders, how many boys are there?

PilotLPTM [1.2K]

Answer:

boys = 250

girls = 150

Step-by-step explanation:

5g = 3b eq. 1

g + b = 400 eq. 2

g = girls

b = boys

From the eq. 2

g = 400 - b

Replacing this last eq. on eq. 1:

5(400-b) = 3b

5*400 + 5*-b = 3b

2000 - 5b = 3b

2000 = 3b + 5b

2000 = 8b

2000/8 = b

250 = b

From eq. 2

g + 250 = 400

g = 400 - 250

g = 150

Check:

from eq. 1

5*150 = 3*250 = 750

3 0

2 years ago

Please help me its important

Amiraneli [1.4K]

Answer:

1. 16

Step-by-step explanation:

first one i believe is 16 dont need points keep them

6 0

2 years ago

At a production process, the produced items are tested for defects. A defective unit is classified as such with probability 0.9,

Snezhnost [94]

Answer:

We use Baye's theorem: P(A)P(B|A) = P(B)P(A|B)

with (A) being defective and

(B) marked as defective

we have to find P(B) = P(A).P(B|A) + P(¬A)P(B|¬A). .......eq(2)

Since P(A) = 0.1 and P(B|A)=0.9,

P(¬A) = 1 - P(A) = 1 - 0.1 = 0.9

and

P(B|A¬) = 1 - P(¬B|¬A) = 1 - 0.85 = 0.15

put these values in eq(2)

P(B) = (0.1 × 0.9) + (0.9 × 0.15)

= 0.225 put this in eq(1) and solve for P(B)

P(B) = 0.4

6 0

2 years ago

You have a box that is a good size for your tape collection. two rows of tapes will fit in the box. the box is 10 inches wide. e

jeyben [28]

Solution:

Two rows of tapes will fit in the box.

Box wide = 10 inches

Each tape wide = 5/8 inches

10/ (5/8) = (10 * 8)/5

= 16

There are two in each row. So, 32 tapes will fit in the box.

3 0

3 years ago

A single card is drawn from a standard 52-card deck. Let D be the event that the card drawn is a black card and let F be the eve

tester [92]

Answer:

The indicated probability of $P(D \cup F')=\frac{25}{26}$

Step-by-step explanation:

Probability of an event E to be;

P(E) = $\frac{Number of events within E}{Total number of possible outcomes}$

As per the given condition:

Total number of possible outcomes = 52 cards.

Let the event be D and F as follows;

D : Drawn card is a black card

F : Drawn card is a 10 card.

Then,

From the given condition:

P(D) = $\frac{26}{52}$ [Out of 52 cards, 26 were black] ,

P(F) = $\frac{4}{52}$ [Out of 52 cards, there are four 10 cards]

For any two events A and B we always have;

$P(A \cup B) = P(A)+P(B)-P(A \cap B)$

Now, we have to find the indicated probability:

$P(D \cup F')=P(D)+P(F')-P(D \cap F')$ ......[1]

First find the P(F');

P(F') =1-P(F) = $1-\frac{4}{52} =\frac{52-4}{52} =\frac{48}{52}$

Also, to find $P(D \cap F')$ .

We use the formula :

For any event A and B independent variable.

$P(A \cap B) =P(A) \cdot P(B)$

then;

$P(D \cap F') =P(D) \cdot P(F') = \frac{26}{52} \cdot \frac{48}{52} =\frac{24}{52}$

Now, substitute these in [1];

$P(D \cup F')=\frac{26}{52} +\frac{48}{52} -\frac{24}{52}$ = $\frac{26+48-24}{52} =\frac{50}{52} = \frac{25}{26}$

Therefore, the probability of $P(D \cup F')=\frac{25}{26}$

3 0

3 years ago