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3 years ago

The area of Ms. Anderson’s office floor measures 120

Mathematics

2 answers:

Maslowich3 years ago

8 0

Answer:

72 square feet not covered

Step-by-step explanation:

100%-40% = 60%

60% of 120 = 0.6x120 = 72

Reil [10]3 years ago

3 0

Answer:

Step-by-step explanation:

because i know

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How do you work out 4/5 of 25/32 as fraction

Sphinxa [80]

$\frac{4}{5}*\frac{25}{32}=\frac{1}{1}*\frac{5}{8}\to\boxed{\frac{5}{8}}$

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3 years ago

What is -7/2(-7/2x+1)=413/8

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4 years ago

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Mr. Gardener has $300,000 in a deposit account and receives 8 1 3 % interest annually. What annual interest (in dollars) does he

Luba_88 [7]

Answer:

The annual interest he receives is $25000

Step-by-step explanation:

We are given

Mr. Gardener has $300,000 in a deposit account

so,

P=300000

receives 8 1 3 % interest annually

so, r=8 1/3 %

$r=8\frac{1}{3} =\frac{8\times 3+1}{3} =\frac{25}{3}$ %

time =1 year

so, t=1

now, we can use interest formula

$I=\frac{P\times r\times t}{100}$

now, we can plug values

$I=\frac{300000\times \frac{25}{3}\times 1}{100}$

now, we can simplify it

$I=25000$

So,

The annual interest he receives is $25000

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4 years ago

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(Hypothetical.) Suppose five measurements of the volume of a liquid yield the following data (in milliliters).

Digiron [165]

Answer:

The 95% confidence interval for the measurements is [48.106, 53.494].

Step-by-step explanation:

The average M of this sample is

$M=\dfrac{1}{5}\sum_{i=1}^5x_i=\dfrac{1}{5}(52+48+49+52+53)\\\\\\M=\dfrac{254}{5}=50.800$

The standard deviation s of this sample is

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^5(x_i-M)^2}$

$s=\sqrt{\dfrac{1}{4}\cdot [(52-50.8)^2+(48-50.8)^2+(49-50.8)^2+(52-50.8)^2+(53-50.8)^2]}$

$s=\sqrt{\dfrac{1}{4}\cdot [(1.44)+(7.84)+(3.24)+(1.44)+(4.84)]}=\sqrt{\dfrac{18.8}{4}}=\sqrt{4.7}\\\\\\s=2.168$

The degrees of freedom are

$df=n-1=5-1=4$

Then, the critical value of t for a 95% CI and 4 degrees of freedom is t=2.776.

The margin of error of the CI is:

$E=t\cdot s/\sqrt{n}=2.776\cdot 2.17/\sqrt{5}=6.024/2.236=2.694$

Then, the lower and upper bounds of the CI are:

$LL=50.800-2.694=48.106\\\\UL=50.800+2.694=53.494$

The 95% confidence interval for the measurements is [48.106, 53.494].

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4 years ago

Find all zeroes of the function. Enter the real zeroes separated by commas.

Elena L [17]

Answer:

f(x)

=(x-4)[6(x^2)+23x+20]

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