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3 years ago

Five less than a number is more than twenty

1 answer:

4 0

Y-5=20
the y represents the unknown # that you used to subtract the five from
to find what y is you will need to ad 5 and 20 then y will be 25
5+20=25
25-5=20
y=25

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2 years ago

Log3 of 5 round to nearest thousandth

Semmy [17]

1.465 is the correct answer I believe.

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3 years ago

Determine if this relationship forms a direct variation. Verify your answer.

leva [86]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=k or y=kx

we have

a) for x=1 y=0.50--------> y/x=0.50/1------> 0.50

b) for x=2 y=1--------> y/x=1/2-------> 0.50

c) for x=3 y=1.50--------> y/x=1.50/3-----> 0.50

d) for x=5 y=2.50--------> y/x=2.50/5----> 0.50

the value of k is equal to 0.50

the relationship forms a direct variation.

the equation is

y=0.50*x

Verify

for x=1

y=0.50*(1)------> y=0.50-----> is correct

for x=2

y=0.50*(2)------> y=1.00-----> is correct

for x=3

y=0.50*(3)------> y=1.50-----> is correct

for x=5

y=0.50*(5)------> y=2.50-----> is correct

4 0

2 years ago

Read 2 more answers

Which symbol correctly completes the statement? -7+2__3+(-) >,<, =

OlgaM077 [116]

Answer:

Step-by-step explanation:

Hope that helps!

4 0

3 years ago

A clinical trial was conducted to test the effectiveness of the drug zopiclone for treating insomnia in older subjects. Before t

aleksklad [387]

Answer:

We cannot say that the mean wake time are different before and after the treatment, with 98% certainty. So the zopiclone doesn't appear to be effective.

Step-by-step explanation:

The goal of this analysis is to determine if the mean wake time before the treatment is statistically significant. The question informed us the mean wake time before and after the treatment, the number of subjects and the standard deviation of the sample after treatment. So using the formula, we can calculate the confidence interval as following:

$IC[\mu ; 98\%] = \overline{y} \pm t_{0.99,n-1}\sqrt{\frac{Var(y)}{n}}$

Knowing that $t_{0.99,15} = 2.602$ :

$IC[\mu ; 98\%] = 98.9 \pm 2.602\frac{42.3}{4} \Rightarrow 98.9 \pm 27.516$

$IC[\mu ; 98\%] = [71.387 ; 126,416]$

Note that $102.8 \in [71.384 ; 126.416]$ so we cannot say, with 98% confidence, that the mean wake time before treatment is different than the mean wake time after treatment. So the zopiclone doesn't appear to be effective.

7 0

3 years ago