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3 years ago

PLZ HELP IF YOU ABSOLUTELY KNOW IM TIMED THX WHICH STATEMENT IS FALSE

Mathematics

2 answers:

user100 [1]3 years ago

7 0

Statement D is False

Yanka [14]3 years ago

3 0

Answer:

Step-by-step explanation:

am i too late:(??

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What is the LCM of 24 and 54?

Fiesta28 [93]

Answer: 216

Step-by-step explanation:

The LCM of 24 and 54 is 216. To find the least common multiple of 24 and 54, we need to find the multiples of 24 and 54 (multiples of 24 = 24, 48, 72, 96 . . . . 216; multiples of 54 = 54, 108, 162, 216) and choose the smallest multiple that is exactly divisible by 24 and 54, i.e., 216.

4 0

2 years ago

Read 2 more answers

Charlie paid 660 for 6 shirts and four belts. One shirt costs three times as much as one belt. If Sam bought 5 shirts and 2 suck

Cerrena [4.2K]

Sam would have to pay $510 to buy 5 shirts and 2 suck belts.

Step-by-step explanation:

Let,

Cost of one shirt = x

Cost of one belt = y

According to given statement;

6x+4y = 660 Eqn 1

x = 3y Eqn 2

Putting value of x from Eqn 2 in Eqn 1

$6(3y)+4y=660\\18y+4y=660\\22y=660$

Dividing both sides by 22

$\frac{22y}{22}=\frac{660}{22}\\y=30$

Putting y=30 in Eqn 2

$x=3(30)\\x=90$

One shirt costs $90 and one belt cost $30.

Price of 5 shirts = 90*5 = $450

Price of 2 belts = 30*2 = $60

Total = 450+60 = $510

Sam would have to pay $510 to buy 5 shirts and 2 suck belts.

Keywords: linear equation, addition

Learn more about addition at:

#LearnwithBrainly

6 0

3 years ago

The manager of a music store has kept records of

Marysya12 [62]

Answer:

(a) $P(x\le 3) = 0.75$

(b) $P(x\le 3) = 0.75$

(b) is the same as (a)

(d) $P(x = 1\ or\ 2) = 0.55$

(e) $P(x > 2) = 0.45$

Step-by-step explanation:

Given

$\begin{array}{ccccccc}{CDs} & {1} & {2} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.30} & {0.25} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

Solving (a): Probability of 3 or fewer CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

Solving (b): Probability of at most 3 CDs

Here, we consider:

$\begin{array}{cccc}{CDs} & {1} & {2} & {3} \ \\ {Prob} & {0.30} & {0.25} & {0.20} \ \ \end{array}$

This probability is calculated as:

$P(x\le 3) = P(1) + P(2) + P(3)$

This gives:

$P(x\le 3) = 0.30 + 0.25 + 0.20$

$P(x\le 3) = 0.75$

(b) is the same as (a)

Solving (c): Probability of 5 or more CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {5} & {6\ or\ more}\ \\ {Prob} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x \ge 5) = P(5) + P(6\ or\ more)$

This gives:

$P(x\ge 5) = 0.05 + 0.05$

$P(x \ge 5) = 0.10$

Solving (d): Probability of 1 or 2 CDs

Here, we consider:

$\begin{array}{ccc}{CDs} & {1} & {2} \ \\ {Prob} & {0.30} & {0.25} \ \ \end{array}$

This probability is calculated as:

$P(x = 1\ or\ 2) = P(1) + P(2)$

This gives:

$P(x = 1\ or\ 2) = 0.30 + 0.25$

$P(x = 1\ or\ 2) = 0.55$

Solving (e): Probability of more than 2 CDs

Here, we consider:

$\begin{array}{ccccc}{CDs} & {3} & {4} & {5} & {6\ or\ more}\ \\ {Prob} & {0.20} & {0.15} & {0.05} & {0.05}\ \ \end{array}$

This probability is calculated as:

$P(x > 2) = P(3) + P(4) + P(5) + P(6\ or\ more)$

This gives:

$P(x > 2) = 0.20+ 0.15 + 0.05 + 0.05$

$P(x > 2) = 0.45$

3 0

2 years ago

Can some help me answer this question ASAP

Maru [420]

Answer:

Step-by-step explanation:

v + w //substitute values

-3i + 2 - 4i //combine like terms

-7i +2

4 0

2 years ago

A total of 1 232 students have taken a course in Spanish, 879 have taken a course in French, and 114 have taken a course in Russ

ElenaW [278]

Answer:

$n(S\cap F \cap R)=7$

Step-by-step explanation:

The Universal Set, n(U)=2092

$n(S)=1232\\n(F)=879\\n(R)=114$

$n(S\cap R)=23\\n(S\cap F)=103\\n(F\cap R)=14$

Let the number who take all three subjects, $n(S\cap F \cap R)=x$

Note that in the Venn Diagram, we have subtracted $n(S\cap F \cap R)=x$ from each of the intersection of two sets.

The next step is to determine the number of students who study only each of the courses.

$n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x$

These values are substituted in the second Venn diagram

Adding up all the values

2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]

2092=2085+x

x=2092-2085

x=7

The number of students who have taken courses in all three subjects, $n(S\cap F \cap R)=7$

3 0

3 years ago