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svp [43]
2 years ago
7

A chemist is using 344 milliliters of a solution of acid and water. If 17.2% of the solution is acid, how many milliliters of ac

id are there? Round your answer to the nearest tenth.​
Mathematics
1 answer:
Rzqust [24]2 years ago
4 0
Iworurieoow this is my answer

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Fiesta28 [93]
The correct point of the graph would be B. (2,-5)
8 0
2 years ago
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The average salary for a construction worker is $29,160 per year. A veterinarian makes on average $84,460. If you consider the e
wel
Hey ! 

I am going to explain what I did to get the answer...

I multiplied $29,160 by 30 = $874,800
Then I multiplied $84,460 by 30 = $2,533,800 

Then I subtracted the results = a difference of $1,659,000

So the answer to your question is, $1,659,000 is the difference. 

I hope I was able to help you out :D <span />
5 0
3 years ago
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Hannah is flying a kite with a trying that is 60 feet long. If Patricia is standing directly below the kite and 16 feet away fro
Nataly_w [17]

Answer:

74.5°

Step-by-step explanation:

Hannah, the kite and Patricia form the vertices of a right-angled triangle with the hypotenuse side the length of the string L = 60 feet and adjacent side the distance between Hannah and Patricia = d = 16 feet.

Let the angle between the string and the sand be Ф.

By trigonometric ratios,

cosФ = adjacent/hypotenuse

= d/L

= 16 feet/60 feet

= 0.2667

Ф = cos⁻¹(0.2667)

= 74.53°

≅ 74.5°

So, he angle between the string and the sand is Ф = 74.5°

7 0
3 years ago
Whats the area of a rectangle 17ft 4ft​
Katena32 [7]

Answer:

68

Step-by-step explanation:

Multiply the numbers to get the area

7 0
3 years ago
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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
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