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LuckyWell [14K]
2 years ago
7

Just do 11 or 12 or both please im really tired and it 1:34 in the morning plz help

Mathematics
1 answer:
kumpel [21]2 years ago
3 0

Answer:

11)<em>(-1)</em>

<em>12)1.7</em>

Step-by-step explanation:

11)y+z/x                

 (-3)+(-2)/5

 (-5)/5

<em>    (-1)</em>

12)x-5z/y

   5+5(-2)/(-3)

   5+(-10)/(-3)

   (-5)/(-3)

   1.666

 <em>    1.7</em>

<em></em>

<em>Hope this helps you</em>

<em>If you have any question just ask me</em>

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The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding
melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{236100}{8} = 29512.5

Sum of squares of differences = 3408750

S.D = \sqrt{\frac{3408750}{7}} = 697.82

Population mean, μ = $30,000

Sample mean, \bar{x} = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars} We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975

Now,

t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894

Since,                  

t_{stat} < t_{critical}

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

8 0
3 years ago
Andre cycled a distance 25km in 150 minutes. What was his average speed (km/h)?
FrozenT [24]

Answer:

10 km/h

Step-by-step explanation:

First, convert 150 minutes to hours:

150/60

= 2.5 hours

Then, to find his speed, divide the km by the number of hours:

25/2.5

= 10 km/h

6 0
3 years ago
Read 2 more answers
A large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and
scoundrel [369]

Answer:

-6.134 to +6.134

Step-by-step explanation:

given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution

X is Normal with mean  =6.1 and std dev = 1 unit

We are to determine the range of values containing 70% of the population of x

We know that normal distribution curve is bell shaped symmetrical about the mean.

So to find 70% range we can use 35% on either side of the mean

Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034

Hence corresponding x value is

x=6.1+1.034*1\\x=6.134

i.e. 70% values lie between

-6.134 to +6.134

3 0
3 years ago
Read 2 more answers
Joe runs 9.25 times around a track in 1, 125.803 seconds. If one lap around the track is 502.3 meters,
musickatia [10]

Step-by-step explanation:

Speed = Distance÷Time

so

502.3÷125.803= 3.99275057

6 0
3 years ago
I have two questions?
Artyom0805 [142]
3n/7=24  multiply both sides by 7

3n=168  divide both sides by 3

n=56

...

Personally, learning the material in any way possible is a good thing to me.
4 0
3 years ago
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