Rozlyn invested $2000 in an account that pays 4.25% annual

Mathematics

1 answer:

densk [106]2 years ago

4 0

9514 1404 393

Answer:

$850

Step-by-step explanation:

The amount of interest earned is ...

I = Prt . . . . . on principal P at annual rate r for t years

Using the given values, the interest earned is ...

I = $2000·0.0425·10 = $850

The interest Rozlyn's account earns in 10 years is $850.

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Temka [501]

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brainly.com/question/18829057

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2 years ago

Robert was using a bucket that holds 243.5 ounces (oz) of water when full. The bucket loses 0.3 oz of water per second. After 20

Assoli18 [71]

Answer:

237.5

Step-by-step explanation:

Since the bucket loses 0.3 oz of water per second, and 20 secs passed, you would have to multiply 0.3 and 20 to find out how much water leaked. And 0.3x20=6, so then, we would have to subtract 234.5 and 6 to find out how much water there is left in the bucket. 234.5-6=237.5

So 237.5 is your answer

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2 years ago

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:

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

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