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2 years ago

after 24.0 days, 2.00 milligrams of an original 128.0 milligram sample remain. what is the half-life of the sample?

1 answer:

6 0

<span>(1/2)n = 0.015625
n log 0.5 = log 0.015625
n = log 0.5 / log 0.015625
n = 6</span><span>24 days / 6 half-lives = 4.00 days (the length of the half-life)

</span>

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12700 = 23400 x (1-11.5/100)n

xxTIMURxx [149]

Answer:

Step-by-step explanation:

889

6 0

2 years ago

Which phrase best describes the translation from the graph y = (x – 5)2 + 7 to the graph of y = (x + 1)2 – 2? A. 6 units left an

VikaD [51]

Answer:

A. 6 units left and 9 units down

Step-by-step explanation:

Given:

y = (x – 5)^2 + 7

y = (x +1)^2 - 2

The graph of y = (x – 5)^2 + 7 is a parabola 5 units right and 7 units up

The graph of y = (x +1)^2 - 2 is a parabola 1 unit left and 2 units down

Transformation is the difference in the shift

Transformation = 6 units left and 9 units down.

Therefore, answer A. 6 units left and 9 units down

Hope this will helpful.

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4 0

2 years ago

Read 2 more answers

A very weak university is on and off probationary status with the accrediting agency. A different procedure is used in July and

tino4ka555 [31]

Answer:

0.3425 = 34.25% probability it will be off probation in February 2020

Step-by-step explanation:

We have these desired outcomes:

Off probation in July 2019, with 0.25 probability, then continuing off in January, with 1 - 0.08 = 0.92 probability.

Still in probation in July 2019, with 1 - 0.25 = 0.75 probability, then coming off in January, with 0.15 probability.

What is the probability it will be off probation in February 2020?

$p = 0.25*0.92 + 0.75*0.15 = 0.3425$

0.3425 = 34.25% probability it will be off probation in February 2020

6 0

2 years ago

From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi

Lana71 [14]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

$P=\frac{45}{70}= 0.64$

$(1-P)=1-0.64=0.36$

$n= 70$

$P$ ± $z$ $\sqrt{\frac{P(1-P)}{n} }$

where the value $z$ will be taken from the z-table for 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table the value of $z$ corresponding to 0.0975 is 1.96

$0.64$ ± $1.96$ $\sqrt{\frac{0.64*0.36}{70} }$

$0.64$ ± $1.96$ $(0.057)$

$0.64$ ± $0.112$

$64%$ % ± $11.2$ %

so the confidence interval is

$64+11.2=75.2$ %

$64-11.2=52.8$ %

$[52.8, 75.2]$

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2 years ago

WILL GIVE BRAINLIEST!!!!!! 50 PTS!!!!!!!!

Softa [21]

Let the original price be p.

Then the sale price was (1.00-0.18)p, or 0.82p.

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Thus, the final cost of the game was 0.82p + 0.0656p = 0.886p.

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2 years ago