Answer: y= Include no spaces in your answer.

Ksenya-84 [330]

My guess would be guess A

3 0

3 years ago

I need assistance solving these questions using Cramers' Rule (which I am frankly terrible at) and need these to finish my homew

Scrat [10]

Answer:

What are the questions?

3 0

3 years ago

Read 2 more answers

Enter the giveaway for points

Nookie1986 [14]

Answer:

Thxx for points how do I enter

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

8. A map has a scale factor of 0.25 inch : 50 miles. On the map, the distance of a river is 2.5 inches. What is the

Arturiano [62]

Set up a ratio for this problem:

.25 in/50 miles = 2.5 in/x

.25x = 125

then divide by .25

The actual distance is: 500 miles

7 0

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:

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

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

$P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\$

$P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886$

7 0

2 years ago