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liq [111]
1 year ago
14

What is the product of StartFraction 4 Over 5 EndFraction times two-fifths? A model representing StartFraction 4 Over 5 EndFract

ion and two-fifths StartFraction 6 Over 25 EndFraction StartFraction 8 Over 25 EndFraction StartFraction 6 Over 10 EndFraction StartFraction 20 Over 10 EndFraction
Mathematics
2 answers:
Anuta_ua [19.1K]1 year ago
3 0

Answer:

StartFraction 8 Over 25

Step-by-step explanation:

What is the product of StartFraction 4 Over 5 EndFraction times two-fifths?

Mathematically =

StartFraction 4 Over 5 EndFraction = 4/5

Two- fifths = 2/5

Hence,

The product of StartFraction 4 Over 5 EndFraction times two-fifths

= 4/5 × 2/5

= 8/25

The correct option =

StartFraction 8 Over 25 EndFraction

coldgirl [10]1 year ago
3 0

Answer:

The answer is 8/25

Step-by-step explanation:

Hope this helps! good luck !

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A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take
ddd [48]

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

7 0
1 year ago
One morning, eight buses arrive at a bus stop.
Semenov [28]

Answer:

0 and 5 minutes

Step-by-step explanation:

Median is the middle observation of given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of middle two value if the total number of observation is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

We have given 8 numbers and we have to find the other two numbers a minute late for each bus to arrive at the bus stop.

Also, we have given the median and mode of all 10 buses i.e. 3.5 and 0 respectively.

Since the mode is 0, so the frequency of 0 must be greater than 2 times and less than or equal to 4 times.

The frequency of 0 is 4 is not possible as then we didn't get a median 3.5 then.

So the frequency of 0 is 3.

Hence one of number is 0 in two unknown number of a minute late for each bus arrive at the bus stop.

Now, we have the number of minutes late for each bus after arranging them in ascending order is: 0 0 0 2 2 6 8 8 9

For getting a median 3.5 we must have a number between 2 and 6. Let it be x then:

\frac{2+x}{2} = 3.5

⇒x = 5

Hence the number of minutes late for each bus is: 0 0 0 2 2 5 6 8 8 9

Thus, 0 and 5 are minutes late for the two-afternoon buses.

6 0
2 years ago
Read 2 more answers
Osama starts with a population of 1,000 amoebas that increases 30% in size every hour for a number of hours, h. The expression 1
Blizzard [7]

Answer: The correct option is A, itis the product of the initial population and the growth factor after h hours.

Explanation:

From the given information,

Initial population = 1000

Increasing rate or growth rate = 30% every hour.

No of population increase in every hour is,

1000\times \frac{30}{100} =1000\times 0.3

Total population after h hours is,

1000(1+0.3)^h

It is in the form of,

P(t)=P_0(t)(1+r)^t

Where P_0(t) is the initial population, r is increasing rate, t is time and [tex(1+r)^t[/tex]  is the growth factor after time t.

In the above equation 1000 is the initial population and (1+0.3)^h is the growth factor after h hours. So the equation is product of of the initial population and the growth factor after h hours.

Therefore, the correct option is A, itis the product of the initial population and the growth factor after h hours.

3 0
1 year ago
Read 2 more answers
A uniform bar of length l and weight w is attached to a wall with a hinge that exerts a horizontal force hx and a vertical force
maria [59]
The answer is Hx = ½ Wsin θ cos θ
The explanation for this is:
Analyzing the torques on the bar, with the hinge at the axis of rotation, the formula would be: ∑T = LT – (L/2 sin θ) W = 0
So, T = 1/2 W sin θ. Analyzing the force on the bar, we have: ∑fx = Hx – T cos θ = 0Then put T into the equation, we get:∑T = LT – (L/2 sin θ) W = 0
4 0
2 years ago
Which equations represent circles that have a diameter of 12 units and a center that lies on the y-axis? Select two options. x2
olga55 [171]

Answer:

x^2 + (y – 3)^2 = 36

Step-by-step explanation:

The standard equation of a circle with center at (h, k) and radius r is

(x - h)^2 + (y - k)^2 = r^2.

If the center lies on the y-axis, then h = 0:  (x - 0)^2 + (y - k)^2 = r^2

If the circle diameter is 12, then the circle radius is 6, and so r^2 = 36

So, among the given equations, your

x^2 + (y – 3)^2 = 36 is correct (but only if you use " ^ " for exponentiation).

3 0
1 year ago
Read 2 more answers
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