Eight times the reciprocal of a number equals 2 times the reciprocal of 9. Find the number

mihalych1998 [28]

So,

To what percent 34 is of 55, we first need to find the fraction equivalent.

34 out of 55 parts or 34/55

Divide 34 by 55.

34/55 = 0.61818...

Now, we move the decimal point 2 places to the right in order to convert it to a percent.

6.1818...
61.81818...

Rounded to the nearest hundredth: 61.82%
Rounded to the nearest tenth: 61.8%
Rounded to the nearest percent: 62%

4 0

2 years ago

Marcie bought a total of 20 used books and cds during a yard sale for a total of 54.50$. of books cost 1.50$ each and cds 5$ eac

melomori [17]

Let numbers of books be 'b' and numbers of CDs be 'c'

We can set up two equations:
Equation [1] ⇒ $b+c=20$
Equation [2] ⇒ $1.50b+5c=54.50$

We are solving for the number of books and the number of CDs bought

When we have two equations in terms of two different variables; $b$ and $c$ , that we need to solve, then this becomes a simultaneous equation problem.

First, rearrange Equation [1] to make either $b$ or $c$ the subject:
$b+c=20$
$b=20-c$

Then we substitute $b=20-c$ into Equation [2]
$1.50b+5c=54.50$
$1.50(20-c)+5c=54.50$
$30-1.50c+5c=54.50$
$5c-1.5c=54.50-30$
$3.5c=24.50$
$c=7$

Now we know the value of $c$ which is $c=7$ , substitute this value into $b=20-c$ we have $b=20-7=13$

Answer:
Numbers of books = 13
Numbers of CDs = 7

5 0

2 years ago

The experimental probability that Kevin will catch a fly ball is equal to 7/8. About what percent of the time will Kevin catch a

Paladinen [302]

Answer:

87.5% percent chance.

Trust me it's right.

6 0

2 years ago

In a completely randomized experimental design involving three assembly methods, 30 employees were randomly selected and 10 were

AnnZ [28]

Answer:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have $3$ groups and on each group from $j=1,\dots,10$ we have $10$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

Where SST represent the total sum of squares.

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=30-3=27$ .

And the total degrees of freedom would be $df=N-1=30 -1 =29$

From the info given we know that $MSR=\frac{SSR}{2}=45.89$

And $MSE=\frac{SSE}{27}=6.27$

From definition the F statisitc is defined as:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

3 0

2 years ago

Louden County Wildlife Conservancy counts butterflies each year. Data over the last three years regarding four types of butterfl

MrRa [10]

Incomple question. However, here's the remaining part of the question:

14

2009

Meadow Fritillary= 5

Variegated Fritillary= 7

Zebra Swallowtail= 33

Eastern-Tailed Blue= 242

Louden County Butterfly Count

2010

Meadow Fritillary 34

Variegated Fritillary 95

Zebra Swallowtail 21

Eastern-Tailed Blue 168

2011

Meadow Fritillary

Variegated Fritillary

Zebra Swallowtail

Eastern-Tailed Blue

10

170

<u>Options</u>:

A) All butterfly populations are steadily decreasing.

B)All butterfly populations were larger than usual in 2010.

C)The Eastern-Tailed Blue butterfly is more common than the others.

D)The Meadow Fritillary is equally common as the Variegated Fritillary

Answer:

<u>C</u>

Step-by-step explanation:

Looking through the above count data by Louden County Wildlife Conservancy from 2009 to 2011 we notice the Eastern- Trailed Blue butterfly has a higher count, which implies that the Eastern-Tailed Blue butterfly is more common than the other butterflies.

Therefore, we could infer from the samples, that the Eastern-Tailed Blue butterfly is more common than others from the records of the past 3 three years.

5 0

2 years ago