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inysia [295]

Answer:

$\displaystyle a_8=142,11\ mg$

Step-by-step explanation:

Geometric sequence

Each term in a geometric sequence can be computed as the previous term by a constant number called the common ratio. The formula to get the term n is

$\displaystyle a_n=a_1r^{n-1}$

where $a_1$ is the first term of the sequence

The problem describes Georgie took 275 mg of the medicine for her cold in the first hour and that in each subsequent hour, the amount of medicine in her body is 91% (0.91) of the amount from the previous hour. It can be written as

amount in hour n = amount in hour n-1 * 0.91

a)

This information provides the necessary data to write the general term as

$\displaystyle a_n=275\ .\ 0.91^{n-1}$

b)

In the 8th hour (n=8), the remaining medicine present is Georgie's body is

$\displaystyle a_n=275\ .\ 0.91^{8-1}$

$\displaystyle a_n=275\ .\ 0.91^{7}$

$\boxed{\displaystyle a_8=142,11\ mg}$

6 0

3 years ago

he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w

stellarik [79]

Answer:

90% confidence interval is ( -149.114, -62.666 )

Step-by-step explanation:

Given the data in the question;

Sample 1 Sample 2

x"₁ = 259.23 x"₂ = 365.12

s₁ = 34.713 s₂ = 48.297

n₁ = 5 n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

$S_p^2$ = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

$S_p^2$ = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

$S_p^2$ = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [ 4 + 9 ]

$S_p^2$ = [ 4819.9692 + 20993.4 ] / [ 13 ]

$S_p^2$ = 25813.3692 / 13

$S_p^2$ = 1985.64378

Now the Standard Error will be;

$S_{x1-x2$ = √[ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

$S_{x1-x2$ = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

$S_{x1-x2$ = √[ 397.128756 + 198.564378 ]

$S_{x1-x2$ = √595.693134

$S_{x1-x2$ = 24.4068

Critical Value = $t_{\frac{\alpha }{2}, df$ = $t_{0.05, df=13$ = 1.771 { t-table }

So,

Margin of Error E = $t_{\frac{\alpha }{2}, df$ × [ ( $S_p^2$ / n₁ ) + ( $S_p^2$ / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666 )

3 0

3 years ago

A football player completes a pass 69.4% of the time. Find the probability that (a) the first pass he completes is the second

wlad13 [49]

Step-by-step explanation:

(a) If his second pass is the first that he completes, that means he doesn't complete his first pass.

P = P(not first) × P(second)

P = (1 − 0.694) (0.694)

P ≈ 0.212

(b) This time we're looking for the probability that he doesn't complete the first but does complete the second, or completes the first and not the second.

P = P(not first) × P(second) + P(first) × P(not second)

P = (1 − 0.694) (0.694) + (0.694) (1 − 0.694)

P ≈ 0.425

(c) Finally, we want the probability he doesn't complete either pass.

P = P(not first) × P(not second)

P = (1 − 0.694) (1 − 0.694)

P ≈ 0.094

6 0

3 years ago

Kent needs to buy school supplies. Notebooks cost 5x dollars, bottles of glue cost 2x dollars packs of markers cost 7x dollars

Damm [24]

Answer: