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matrenka [14]
3 years ago
8

When Arjun makes hot cocoa, he adds 20\text{ mL}20 mL20, start text, space, m, L, end text of chocolate syrup for every 333 ounc

es of hot water. Arjun's brother, Eli, adds 45\text{ mL}45 mL45, start text, space, m, L, end text of chocolate syrup for every 666 ounces of hot water. Which hot cocoa is more chocolatey? Choose 1 answer: Choose 1 answer:
Mathematics
1 answer:
loris [4]3 years ago
7 0

Answer:

<h3>45mL hot cocoa is more chocolatey</h3>

Step-by-step explanation:

When Arjun makes hot cocoa, if he add 20mL of chocolate syrup to 3ounces of water, let us know the equivalent chocolate in one ounces of water.

20mL = 3 ounces

x = 1 ounce

cross multiply

3x = 20

x = 20/3

x = 6.67 mL

This means he must have add 6.67mL of chocolate to 1 ounce of water.

Similarly, if 45mL is added to every 6 ounces

45mL = 6ounces

x = 1 ounce

cross multiply

6x = 45

x = 45/6

x =  7.5mL

This means he must have add 7.5mL of chocolate to 1 ounce of water.

The hot cocoa that is more chocolatey is the one with the highest amount of chocolate in 1 ounce. That is 45mL hot cocoa

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Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
A jar contains 18 apple lollipops, 7 strawberry lollipops, and 10 grape lollipops. Irvin randomly selects a lollipop, gives it t
andrey2020 [161]
First  choice is  "1 of 10 grape lollipops from 35 lollipops"  
The probablity is 10/35 = 2/7 
Second choice is  "1 of 18 apple lollipops from 34 lollipops"  
So probablity is 18/34 = 9/17 
Between this is  AND so you have to these probablities multiply:
P(A) = 2/7 * 9/17 = 18/119 
P(A) = 18/119 * 100% = 1800/119 %
It is approximaly 15,1 %. 
4 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
inysia [295]

\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

5 0
2 years ago
Given: Triangle ABC, mmBC=24 cm, line segment AL- &lt; bisector
Furkat [3]

The value of line AL is 21. 51cm

<h3>How to determine the length</h3>

To find line AL,

Using

Sin α = opposite/ hypotenuse to find line AB

Sin 90 = x/ 24

1 = x/24

Cross multiply

x = 24cm

Now, let's find line AC

Sin angle B = line AC/24

Note that to find angle B

angle A + angle B + angle C = 180

But angle B = 2 Angle A

x + 2x + 90 = 180

3x + 90 = 180

3x = 180-90

x = 30°

Angle B = 2 × 30 = 60°

Sin 60 = x/ 24

0. 8660 = x/24

Cross multiply

x = 24 × 0. 8660

x = 20. 78cm

We have the angle of A  in the given triangle to be divide into two by the bisector, angle A = 15°

To find line AL, we use

Cos = adjacent/ line AL

Cos 15 = 20. 78/ line AL

Line AL = 20. 78/ cos 15

Line AL = 20. 78 / 0. 9659

Line AL = 21. 51 cm

Thus, the value of line AL is 21. 51cm

Learn more about trigonometry ratio here:

brainly.com/question/24349828

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6 0
1 year ago
Which pairs of faces will be opposite one another when the prism is created? Check all that apply.
blondinia [14]
A&C
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Try it by fold the prism with paper
7 0
3 years ago
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