PLEASE it’s due tonight help me! All help will be greatly appreciated.

Miranda wants to give her 14-year-old daughter $20,000 when she turns 18. How much does she need to put in the bank now if the i

Eddi Din [679]

20,000 = P ( 1 + 0.1 )^4
20,000 = P * 1.1^4
20,000 = P * 1.461
P = 20,000 : 1.461 ≈ $13,689.25
Answer: Miranda needs to put now in the bank $13,689.25

3 0

2 years ago

Mrs. CADY CONSTRUCTS A CUBE WITH 512 MAGNETIC BLOCKS. STUDENTS IN HER TWO CLASSES WILL EACH MAKE AN IDENTICAL CUBE. THERE ARE 28

Lilit [14]

YOU DON"T HAVE TO YELL

512 PER STUDENT

BASICALLY,
CUBES NEEDED=CUBES PER STUDENT TIMES NUMBER OF STUDENTS

NUMBER OF STUDENTS=28+25=53

CUBES PER STUDENT=512

CUBES NEEDED=512 TIMES 53 EQUALS 27136 CUBES

SHE NEEDS 27136 CUBES FOR ALL HER STUDENTS

8 0

2 years ago

if you multiply the number by 2 and add 4, then result you get will be the same as three times the number decreased by 7.

liraira [26]

Answer:

Step-by-step explanation:

Let the number be x,

2x+4 = 3x-7

Collect like terms

2x-3x= -4-7

-x =. -11

Multiply both sides by -1

x. =. 11

PLEASE MARK BRAINLIEST.

5 0

2 years ago

Memory module consists of 9 chips. The device is designed with redundancy so that it works even if one of its chips is defective

soldier1979 [14.2K]

Answer:

a) $P[C]=p^n$

b) $P[M]=p^{8n}(9-8p^n)$

c) n=62

d) n=138

Step-by-step explanation:

Note: "Each chip contains n transistors"

a) A chip needs all n transistor working to function correctly. If p is the probability that a transistor is working ok, then:

$P[C]=p^n$

b) The memory module works with when even one of the chips is defective. It means it works either if 8 chips or 9 chips are ok. The probability of the chips failing is independent of each other.

We can calculate this as a binomial distribution problem, with n=9 and k≥8:

$P[M]=P[C_9]+P[C_8]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)\\\\P[M]=p^{8n}(p^{n}+9(1-p^n))\\\\P[M]=p^{8n}(9-8p^n)$

c)

$P[M]=(0.999)^{8n}(9-8(0.999)^n)=0.9$

This equation was solved graphically and the result is that the maximum number of chips to have a reliability of the memory module equal or bigger than 0.9 is 62 transistors per chip. See picture attached.

d) If the memoty module tolerates 2 defective chips:

$P[M]=P[C_9]+P[C_8]+P[C_7]\\\\P[M]=\binom{9}{9}P[C]^9(1-P[C])^0+\binom{9}{8}P[C]^8(1-P[C])^1+\binom{9}{7}P[C]^7(1-P[C])^2\\\\P[M]=P[C]^9+9P[C]^8(1-P[C])+36P[C]^7(1-P[C])^2\\\\P[M]=p^{9n}+9p^{8n}(1-p^n)+36p^{7n}(1-p^n)^2$

We again calculate numerically and graphically and determine that the maximum number of transistor per chip in this conditions is n=138. See graph attached.

6 0

3 years ago

X + 3y = -3 solve for Y (ex. y= 3 + 5x

AnnyKZ [126]

Answer:

y = -1/3x - 1

Step-by-step explanation:

Let's solve for y.

Step 1: Add -x to both sides.

x+3y+−x=−3+−x

3y=−x−3

Step 2: Divide both sides by 3.

y=-1/3x-1

--

$\frac{-1}{3} x-1$

3 0

2 years ago