Chris received $50,000 from his grandparents as a graduation gift. So, he has decided to invest the money

Mathematics

1 answer:

GenaCL600 [577]2 years ago

7 0

Answer: $7142.86

Step-by-step explanation:

Let the amount to be invested in the bond be represented by y.

Based on this, the amount to be invested in the CD will be: = 50000 - y.

From the information given, yearly interest on bond is 12%, this will be:

= 12% × y = 0.12 × y = 0.12y

Also, interest on CD will be:

= 5% × (50000 - y)

= 0.05(50000 - y)

Then, the equation to get the amount to be invested in bond will be:

0.12y + 0.05(50000 - y) = 3000

0.12y + 2500 - 0.05y = 3000

0.12y - 0.05y = 3000 - 2500

0.07y = 500

y = 500/0.07

y = 7142.8571

y = 7142.86

The amount to be invested in bond will be $7142.86

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What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i

Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

Step-by-step explanation:

The question is incomplete:

What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions?

We can model this as a binomial random variable, with p=0.57 and n=14.

$P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}$

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

$P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\$

$P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\$