All categories

3 years ago

1. What is the mean of the data set?

Mathematics

2 answers:

enyata [817]3 years ago

5 0

Answer:

1. mean is 17.4

2. interquartile range is 30

3. mean absolute deviation is 3

4. median is 51

Step-by-step explanation:

aleksandrvk [35]3 years ago

4 0

Answer:

1 = 17.4

4 = 51

You might be interested in

A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th

Yuri [45]

Answer:

$15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

$y(0)=0$ , $y'(0)=0$

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: $F_{r} (t)$ that correspond to the force of resistance on the mass by the action of the spring and $F(t)$ that is an external force with unknown direction (that does not specify in the enounce).

For determinate $F_{r} (t)$ we can use Hooke's Law given by the formula $F_{r} (t) = k y(t)$ where $k$ correspond to the elastic constant of the spring and $y(t)$ correspond to the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

$k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m} = 441.45 \frac{N}{m}$

Now we apply Newton's 2nd Law and obtaint that...

$F_{r} (t)$ ± $F(t)$ = $ma(t)$

$F_{r} (t) = ky(t) = 441.45y(t)$

$F(t) = 170 cos(5t)$

$m = 15 kg$

$a(t) = \frac{d^{2}y(t)}{dt^{2} }$

Finally... $15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) =$ ± $170 cos(5t)$

We know from the problem that there's not initial displacement and initial velocity, so... $y(0)=0$ and $y'(0)=0$

Finally the Initial Value Problem that models the situation describe by the problem is

$\left \{ 15\frac{d^{2}y(t)}{dt^{2} } - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.$

6 0

3 years ago

A 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider.

Zigmanuir [339]

Answer: $(20.0\%,24.0\%)$

Step-by-step explanation:

Given, A 2015 Gallup poll of 1,627 adults found that only 22% felt fully engaged with their mortgage provider.

Here , 1,627 adults are determining the sample.

Thus, sample size : n = 1627

Also, the sample proportion of adults felt fully engaged with their mortgage provider are $\hat{p}=22\%=0.22$

for 95% confidence level , critical z-value =1.96

Then , the 95% confidence interval would be :-

$\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Substituting values , we get

$0.22\pm 1.96(\sqrt{\dfrac{0.22(1-0.22)}{1627}})\\\\=0.22\pm 1.96\sqrt{\dfrac{0.22\times0.78}{1627}}\\\\\approx0.22\pm0.0201\\\\=(0.22-0.0201,\ 0.22+0.0201)\\\\=(0.1999,\ 0.2401)\approx(20.0\%,24.0\%)$

Hence, the required 95% confidence interval using only 1 decimal : $(20.0\%,24.0\%)$ .

4 0

3 years ago

What is the correct domain of h

Olenka [21]

the correct domain oh h is C

6 0

3 years ago

A bag has ten marbles in it.Three of them are green one of them is red four of them are yellow and two of them are blue.If one m

yanalaym [24]

Since there’s 3 green and 4 yellow that adds up to 7 and with only 3 marbles left that is 3:10 and that make the probability 30% or 0.30

7 0

3 years ago

Read 2 more answers

Which of these are written in scientific notation

SOVA2 [1]

A properly written number in scientific notation has the form:
+/- A*10^n
where +/- is either plus or minus
A is a decimal number (or integer) where 1<= A < 10
and n is a positive or negative integer.

According to these rules,
120,000,000 is written in scientific notation as
1.2 * 10^8
so we need to move the decimal point 8 places to the left and multiply by 10^8.

4 0

3 years ago