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3 years ago

9

Nemo and Dory are meeting Crush at the tide pool. Nemo must swim 6 ¾ miles while Dory must swim 4 1/2 miles to reach their desti

nation. How many times farther does Nemo have to swim?

1 answer:

gayaneshka [121]3 years ago

5 0

Answer:

1.5 times.

Step-by-step explanation:

6.75 / 4.5 = 1.5.

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Can u please help my friend with this she really needs it

Gnoma [55]

Answer:

D

Step-by-step explanation:

The answer is D

3 0

3 years ago

Speedy Swift is a package delivery service that serves the greater Atlanta, Georgia, metropolitan area. To maintain customer loy

Paladinen [302]

Answer:

See Explanation and attachments

Step-by-step explanation:

Given

See question for data

Solving (a):

The scale used is an ordinal scale.

Ordinals scale uses hierarchy arrangement and the data can be arranged as:

$EARLY ->> ON-TIME ->> LATE ->> LOST$

The variable of the delivery performance is qualitative because it is non-numerical

Solving (b): Frequency table

From the table, we have:

$On - Time = 57$ $Early = 19$ $Late = 9$ $Lost = 2$

So, the frequency distribution table is:

$\begin{array}{cc}{Performance}&{Frequency}&{On-Time}&{57}&{Early}&{19}&{Late}&{9} & {Loss} & {2} & {Total} & {87}\end{array}$

Solving (c): Frequency table.

To do this, we add another column (Relative Frequency) to the above table.

The relative frequency is calculated as:

$Relative\ Frequency = \frac{Frequency}{Total}$

So, we have:

$On - Time = \frac{57}{87} = 0.656$ $Early = \frac{19}{87} = 0.218$

$Late = \frac{9}{87} = 0.103$ $Lost = \frac{2}{87} = 0.023$

So, the frequency distribution table is:

$\begin{array}{ccc}{Performance}&{Frequency}&{Relative\ Frequency} & {On-Time}&{57}&{0.656}&{Early}&{19}&{0.218} & {Late}&{9} & {0.103} & {Loss} & {2} & {0.023} & {Total} & {87}&{1}\end{array}$

Solving (d & e): See attachment 1 for bar chart & attachment 2 for pie chart

Solving (f):

From the question, we understand that the object is to return 99% early or on time and never to lose a package.

The analysis is as follows:

Early and On-Time (ET) packages

$ET = \frac{Early + On-Time}{Total} * 100\%$

$ET = \frac{57 + 19}{87} * 100\%$

$ET = \frac{76}{87} * 100\%$

$ET = \frac{7600}{87} \%$

$ET = 87.4 \%$

Lost packages

$Lost = \frac{Lost}{Total} * 100\%$

$Lost = \frac{2}{87} * 100\%$

$Lost = \frac{200}{87} \%$

$Lost = 2.30\%$

<em>From the above analysis, we can see that 87.4% of the packages were delivered early enough and 2.30% were lost. </em>

<em>The fraction of packages delivered early can be improved and the fraction of lost packages can be reduced by exploring the chances of taking alternative routes when possible. </em>

3 0

3 years ago

Write sin 64° in terms of cosine.

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

Rule

The sin of any angle = cos(90 - the angle used for the sine)

Sin(64) = cos(90 - 64)

Sin(64) = cos(26)

Check

Sin(64) = 0.8988

Cos(26) = 0.8988

6 0

3 years ago

Read 2 more answers

2(x-5)=10 but the pemdas is backwards

ale4655 [162]

Answer:

2 (x-5) = 10

2x-10 = 10

2x = 10 + 10

x = 20/2

x = 10

Step-by-step explanation:

8 0

3 years ago

Use the remainder theorem to determine the remainder when 3t²+5t-7 is divided by t-5

Brrunno [24]

$f(t)=3t^2+5t-7\implies f(5)=3(5)^2+5(5)-7=93$

which means the remainder upon dividing $f(t)$ by $t-5$ is 93.

This is because we can write

$\dfrac{f(t)}{t-5}=q(t)+\dfrac{r(t)}{t-5}$

where $q(t)$ is the quotient and $r(t)$ is the remainder. Multiplying both sides by $t-5$ , then evaluating at $t=5$ makes the $q(t)$ term vanish, leaving the remainder:

$f(t)=q(t)(t-5)+r(t)\implies f(5)=r(5)$

6 0

4 years ago