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3 years ago

Holly missed 6 of 181 school days this year. Approximately what percent of school days was Holly in attendance?

Mathematics

1 answer:

Natasha2012 [34]3 years ago

4 0

$181-6=175\\\\\frac{175}{181}\cdot100\%=\\\\\frac{17500}{181}\%=\\\\96.7\%\approx97\%\\\\Answer\ C$

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A company that produces computers recently tested the battery in its latest laptop in six separate trials. The battery lasted 8.

Anna35 [415]

confidence interval for the mean battery life in the new model is [7.9489, 8.3045].

What is confidence interval ?

A confidence interval, in statistics, refers to the chance that a population parameter can fall between a collection of values for an exact proportion of times.

Main body:

Formula for confidence interval is =

CI = x- bar ± z*s/√n where,

CI = confidence interval

x- bar = sample mean

z = confidence level value

{s} = sample standard deviation

{n} = sample size

given ;

n = 6

mean = (8.23+7.89+8.14+8.25+8.30+ 7.95)/6

= 8.127

value of z for 95% C.I. = 1.96

C.I. = 8.127 ± 1.96 * 0.22/√6

C.I. = 8.127 ± 0.1781

C.I. =[7.9490, 8.3044]

Hence correct option is A.

To know more about confidence interval , visit:;

brainly.com/question/15712887

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8 0

11 months ago

What do u have 2 do to solve

Nastasia [14]

5x^2 + 6x +13.88 is what is is simplified hope it helped a bit

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3 years ago

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The length of AB¯¯¯¯¯ is 15 centimeters. A dilation with a scale factor of 0.4 is applied. What is the length of the image of

Vitek1552 [10]

The answer is 6 hope this helps

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3 years ago

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The population of California was 29.76 million in 1990 and 33.87 million in 2000. Use the model found in part a (f(x) =29.76(1.0

NeTakaya

Answer:

The average rate of change from 1990 to 1991 is $0.387 \:\frac{million}{years}$

The average rate of change from 2000 to 2001 is $0.440 \:\frac{million}{years}$

Step-by-step explanation:

The average rate of change of function f(x) over the interval $a\leq x\leq b$ is given by

$\frac{f(b)-f(a)}{b-a}$

It is a measure of how much the function changed per unit, on average, over that interval.

From the information given:

The function that models the population t years after 1990 is $f(t) =29.76(1.013)^t$
The year 1990 is t = 0 and the year 2000 is t = 10.

1. The average rate of change from 1990 to 1991 is:

The interval is $0\leq x\leq 1$

$\frac{29.76(1.013)^1-29.76(1.013)^0}{1-0}\\\\\frac{1.013^1\cdot \:29.76-1\cdot \:29.76}{1-0}\\\\\frac{0.38688}{1-0}\\\\0.387 \:\frac{million}{years}$

2. The average rate of change from 2000 to 2001 is

The interval is $10\leq x\leq 11$

$\frac{29.76(1.013)^{11}-29.76(1.013)^{10}}{11-10}\\\\\frac{0.44022}{11-10}\\\\0.440 \:\frac{million}{years}$

8 0

3 years ago

Pls help it would be greate if you could answer all thx with show all work

ira [324]

Problem 1:

1. distribute the 4 to (x-3)

2. 2x-6+4x-12

3. Combine like terms; 6x-18

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1. Combine the equations; 10w+120+45w+25

2. Combine like terms; 55w+ 145

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1. Distribute the - to the g+11

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3. -3g-4

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3 years ago