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Margarita [4]
3 years ago
14

How can u Simplify negative 9 over 6 divided by 3 over negative 2. .

Mathematics
1 answer:
kirill115 [55]3 years ago
8 0

Answer:

negative 3 over 2 divided by 3 over negative 2.

explanation:

negative 9 over 6 simplified is:

3:2 as a ratio which is negative 3 over 2.

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At the theater, tickets for adults cost $4.75
shutvik [7]
X+y= 26 and 4.75x + 2.25y= 83.50

y=26-x( substitute this into second equation for y and solve for x)

4.75x + 2.25(26-x)= 83.50

4.75x + 58.5-2.25x= 83.50

x= 10

Now solve for y by substituting your answer for x

10+ y= 26

y=16

Therefore, 16 tickets were purchased for kids and 10 for adults.

7 0
2 years ago
What is the modulus of 2 – 3i?
Bas_tet [7]
What is the modulus of 2 – 3i?

8 0
3 years ago
A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying techn
tester [92]

Answer:

a) We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

b) z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

c) p_v =2*P(z  

d) Null hypothesis:p=0.35  

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

\hat p=\frac{522}{1562}=0.334 estimated proportion of people who have heard of a new electronic reader

p_o=0.35 is the value to verify

\alpha=0.05 represent the significance level

z would represent the statistic

p_v represent the p value

Part a

We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

Part b

The statistic for this case is given :

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

p_v =2*P(z  

Part d

The null hypothesis for this case would be:

Null hypothesis:p=0.35  

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

5 0
3 years ago
How would you do a,b,c,and d
I am Lyosha [343]

a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as

... c = k·s² + m·s + n

Filling in the given values gives three equations in k, m, and n.

28 = k\cdot 10^2+m\cdot 10+n\\21=k\cdot 20^2+m\cdot 20+n\\16=k\cdot 30^2+m\cdot 30+n

Subtracting each equation from the one after gives

-7=300k+10m\\-5=500k+10m

Subtracting the first of these equations from the second gives

2=200k\\\\k=\dfrac{2}{200}=0.01

Using the next previous equation, we can find m.

-5=500\cdot 0.01+10m\\\\m=\dfrac{-10}{10}=-1

Then from the first equation

[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]

There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with

... c = 0.01s² - s + 37

b) At 150 kph, the cost is predicted to be

... c = 0.01·150² -150 +37 = 112 . . . cents/km

c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.

d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.

4 0
3 years ago
the speed limit on a highway is 110 km per hour. how much time does it take a car to travel 132 km at this speed
satela [25.4K]

Given:

Speed limit on a highway is 110 km per hour.

Let x be the speed of the car in time 't'

t\ge\frac{x}{110}

time when taken for a car to travel 132km.

\begin{gathered} t\ge\frac{132}{110} \\ t\ge1.2\text{ hours} \end{gathered}

It takes minimum 1 hour 12 minutes

6 0
1 year ago
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