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Kazeer [188]
2 years ago
12

Connie collected 50 seashells during her 5-day vacation. Each day she collected 3 more than the previous day. How many seashells

did she find each day? PLEASE HELP!
Mathematics
1 answer:
vfiekz [6]2 years ago
6 0

Answer:

The number of seashells collected on day 1 to day 5 is 4, 7, 10, 13 and 16 respectively

Step-by-step explanation:

This is a question that deals with consecutive numbers.

Let x represent the number of seashells collected on her first day.

Given that she collected 3 more each day, then

On day 2, she collected x + 3

On day 3, she collected x + 6

On day 4, she collected x + 9

On day 5, she collected x + 12

Adding up the above will give the number of sea shells collected on day 1;

i.e.

x + x + 3 + x + 6 + x + 9 + x + 12 = 50

Collect like terms

x + x + x + x + x + 3 + 6 + 9 + 12 = 50

5x + 30 = 50

5x = 50 - 30

5x = 20

Multiply ⅕ to both sides

⅕ * 5x = ⅕ * 20

x = 4.

This mean that she found 4 seashells on day 1;

Recall that day 2 = x + 3

So, day 2 = 4 + 3 = 7 seashells

day 3 = x + 6

day 3 = 4 + 6 = 10 seashells

day 4 = x + 9

day 4 = 4 + 9 = 13 seashells

day 5 = x + 12

day 5 = 4 + 13 = 16 seashells

Hence, the number of seashells collected on day 1 to day 5 is 4, 7, 10, 13 and 16 respectively

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3 years ago
A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th
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Answer:  The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by (\mu ):-

\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}

, where n= Sample size.

s= Sample standard deviation

\overline{x} = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

\overline{x}=84

n= 64

Degree of freedom : df = n-1=63  

s= 5.5

Significance level : \alpha=1-0.90=0.1

Two-tailed T-value for df = 63 and  \alpha=1-0.90=0.1 would be

t_{\alpha/2,df}=t_{0.05,63}=1.669  (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

84\pm (1.669)\dfrac{5.5}{\sqrt{64}}

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∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

6 0
3 years ago
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