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2 years ago

Six and one-half ounce is milliliters.

Mathematics

1 answer:

Gennadij [26K]2 years ago

3 0

207.014
I’m pretty sure that’s right :)

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A fraction reduces to 36. If its denominator is 6x5, what is its numerator?

olchik [2.2K]

If you would like to find the numerator of a fraction, you can do this using the following steps:

denominator ... 6 * 5 = 30
numerator ... x = ?

36 = numerator / denominator
36 = x / 30
x = 36 * 30
x = 1080

The correct result would be 1080.

4 0

2 years ago

If a bicyclist traveled at 8:am to 8:18am a distance of 4.5 miles and from 8:18am to 8:45am a distance of 7.5 miles .What is the

Vinil7 [7]

8 am at 0

8:18 at 4.5

8:48 at 7.5 miles ???

18 min = .3 hour

48 min = .8 hour

in first .3 hours speed = 4.5/.3 = 15 mph

in next .5 hours = speed = 3/.5 = 6 mph

average speed for the whole trip = 7.5 miles/.8 hours = 9.375 mph

3 0

2 years ago

The length width and height of a rectangular solid are 8,4,1 respectively what is the length of the longest line segment whose e

LUCKY_DIMON [66]

Answer:

9 units

Step-by-step explanation:

Given

$Length = 8$

$Width = 4$

$Height = 1$

Required

Determine the length of the longest segment

To do this, we use:

$Longest = \sqrt{Length^2 + Width^2 + Height^2}$

So, we have:

$Longest = \sqrt{8^2 + 4^2 + 1^2}$

$Longest = \sqrt{64 + 16 + 1}$

$Longest = \sqrt{81}$

Take positive square root

$Longest = 9$

<em>Hence, the length of the longest segment is 9 units</em>

3 0

2 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$