All categories

3 years ago

Solve 2(3-a)=18 for a

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

8 0

Answer:

Step-by-step explanation:

Alinara [238K]3 years ago

7 0

Answer:

a=-6

Step-by-step explanation:

2(3-a)=18

3-a=18/2

3-a=9

a=3-9

a=-6

You might be interested in

Math help please TY :3 HWAJDSK 10 POINTS REWARDING !!

valkas [14]

I will do Point A carefully, The others I will indicate. Start with the Given Point A. Then do the translations

A(-1,2) Original Point

Reflection: about x axis:x stays the same; y becomes -y:Result(-1,-2)

T<-3,4>: x goes three left, y goes 4 up (-1 - 3, -2 + 4): Result(-4,2)

R90 CCW: Point (x,y) becomes (-y , x ) So (-4,2) becomes(-2, - 4): Result (-2, - 4)

B(4,2) Original Point

Reflection: (4, - 2)
T< (-3,4): (4-3,-2 + 4): (1 , 2)
R90 CCW: (-y,x) = (-2 , 1)

C(4, -5) Original Point

Reflection (4,5)
T<-3,4): (4 - 3, 5 + 4): (1,9)
R90, CCW (-9 , 1)

D(-1 , -5) Original Point

Reflection (-1,5)
T(<-3,4): (-1 - 3, 5 + 4): (-4,9)
R90, CCW ( - 9, - 4)

Note: CCW means Counter Clockwise

The graph on the left is the same one you have been given.

The graph on the right is the same figure after all the transformations

7 0

3 years ago

a machinist can produce 114 Parts in 6 minutes at this rate how many parts can the machinist is produce in 15 minutes

tigry1 [53]

114/6=x/15
6x=1710
x=285 parts

4 0

3 years ago

There are 52 cards in a deck, and 13 of them are hearts. Four cards are dealt, one at a time, off the top of a well-shuffled dec

irga5000 [103]

Answer:

10.97%

Step-by-step explanation:

There are 52 cards.

13 of them, are hearts.

Then

52 - 13 = 39 cards are not hearts.

4 cards are drawn, we want to find the percent chance that the fourth card is a heart card, but no before.

So the first card can't be a heart card.

because the deck is well-shuffled, all the cards have the same probability of being drawn.

Then the probability of not getting a heart card, is equal to the quotient between the number of non-heart cards (39) and the total number of cards (52), then the probability is:

p₁ = 39/52

The second card also can't be a heart card, the probability is calculated in the same way than above, but now there are 38 non-heart cards and a total of 51 cards (because one card was already drawn) then the probability here is:

p₂ = 38/51

For the third card the reasoning is similar to the two above cases, here the probability is:

p₃ = 37/50

The fourth card should be a hearts card, the probability is computed in the same way than above, as the quotient between the number of heart cards in the deck (13) and the total number of cards in the deck (now there are 49 cards)

then the probability is:

p₄ = 13/49

The joint probability (the probability of these 4 events happening together) is equal to the product between the individual probabilities:

P = p₁*p₂*p₃*p₄

P = (39/52)*(38/51)*(37/50)*(13/49) = 0.1097

The percent chance is the above number times 100%

Percent = 0.1097*100% = 10.97%

3 0

3 years ago

A random experiment was conducted where a Person A tossed five coins and recorded the number of ""heads"". Person B rolled two d

cestrela7 [59]

Answer:

(10) Person B

(11) Person B

(12) $P(5\ or\ 6) = 60\%$

(13) Person B

Step-by-step explanation:

Given

Person A $\to$ 5 coins (records the outcome of Heads)

Person $\to$ Rolls 2 dice (recorded the larger number)

Person A

First, we list out the sample space of roll of 5 coins (It is too long, so I added it as an attachment)

Next, we list out all number of heads in each roll (sorted)

$Head = \{5,4,4,4,4,4,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,0\}$

$n(Head) = 32$

Person B

First, we list out the sample space of toss of 2 coins (It is too long, so I added it as an attachment)

Next, we list out the highest in each toss (sorted)

$Dice = \{2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\}$

$n(Dice) = 30$

Question 10: Who is likely to get number 5

From person A list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Head)}$

$Pr(5) = \frac{1}{32}$

$Pr(5) = 0.03125$

From person B list of outcomes, the proportion of 5 is:

$Pr(5) = \frac{n(5)}{n(Dice)}$

$Pr(5) = \frac{8}{30}$

$Pr(5) = 0.267$

From the above calculations:  $0.267 > 0.03125$  Hence, person B is more likely to get 5

Question 11: Person with Higher median

For person A

$Median = \frac{n(Head) + 1}{2}th$

$Median = \frac{32 + 1}{2}th$

$Median = \frac{33}{2}th$

$Median = 16.5th$

This means that the median is the mean of the 16th and the 17th item

So,

$Median = \frac{3+2}{2}$

$Median = \frac{5}{2}$

$Median = 2.5$

For person B

$Median = \frac{n(Dice) + 1}{2}th$

$Median = \frac{30 + 1}{2}th$

$Median = \frac{31}{2}th$

$Median = 15.5th$

This means that the median is the mean of the 15th and the 16th item. So,

$Median = \frac{5+5}{2}$

$Median = \frac{10}{2}$

$Median = 5$

Person B has a greater median of 5

Question 12: Probability that B gets 5 or 6

This is calculated as:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

From the sample space of person B, we have:

$n(5\ or\ 6) =n(5) + n(6)$

$n(5\ or\ 6) =8+10$

$n(5\ or\ 6) = 18$

So, we have:

$P(5\ or\ 6) = \frac{n(5\ or\ 6)}{n(Dice)}$

$P(5\ or\ 6) = \frac{18}{30}$

$P(5\ or\ 6) = 0.60$

$P(5\ or\ 6) = 60\%$

Question 13: Person with higher probability of 3 or more

Person A

$n(3\ or\ more) = 16$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Head)}$

$P(3\ or\ more) = \frac{16}{32}$

$P(3\ or\ more) = 0.50$

$P(3\ or\ more) = 50\%$

Person B

$n(3\ or\ more) = 28$

So:

$P(3\ or\ more) = \frac{n(3\ or\ more)}{n(Dice)}$

$P(3\ or\ more) = \frac{28}{30}$

$P(3\ or\ more) = 0.933$

$P(3\ or\ more) = 93.3\%$

By comparison:

$93.3\% > 50\%$

Hence, person B has a higher probability of 3 or more

7 0

2 years ago

How much water can the cup hold when full?

MatroZZZ [7]

Answer:

92.45cm^2

Step-by-step explanation:

The volume of the cone is found with pi*r^2*h/3

7 0

3 years ago