Whats 354 sqaured ddd

Write 9n + 1 in a word phrase 9n +1

Lisa [10]

Answer:

Sum of 1 and nines times a number

6 0

3 years ago

Use the diagram to answer the following questions

n200080 [17]

Answer:

a. line AR, line AV

b. plane TSR, plane QTZ, plane RSZ

c. A, R, V

d. Q, R, S, T

e. QR

f. RQ, RS

Step-by-step explanation:

a. In addition to a name given to a line ("line p"), a line can be named by any two points on that line. Here, possible names are ...

line AR, line RA, line AV, line VA, line RV, line VR

any two of these names will appropriately answer this part of the question.

__

b. In addition to a name given to a plane ("plane W"), a plane can be named by any three points in that plane. Here, there are 5 named points in the plane (3 of which are collinear), so there are 54 possible different names for the plane. Some of them are ...

plane RST, plane TSR, plane QTZ, plane RSZ

__

c. In this diagram, each of the lines shown has three named points on it. Possible answers to this part of the question are ...

{Q, R, S} or {A, R, V}

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d. Any four of the 5 named points shown in plane W will serve to answer this question. Possible answers are ...

{R, S, T, Z} or {Q, S, T, Z} or {Q, R, T, Z} or {Q, R, S, Z} or {Q, R, S, T}

__

e. Any pair of collinear points will name a line segment. Here, there are 12 possible names for line segments, including ...

QR, RQ, QS, SQ, RS, SR, AR, RA, ...

__

f. Opposite rays can be named by choosing the end point as the middle of three collinear points. Here, the only such middle point is R. We can name pairs of opposite rays on either line. The endpoint of the ray is named first.

{RA, RV} or {RQ, RS}

7 0

3 years ago

Four boys weighed themselves in different pairs and alone. The weights of the pairs were 110 kg, 112 kg, 113 kg, 118 kg, 121 kg,

brilliants [131]

The lost pair of weight can be found by subtracting the a known pair from

the total weight leaving the value of the lost pair.

The lost weight of a pair of boys is 120 kg

Reasons:

Let a, b, c, and d represent the weights of the four boys, we have;

The number of ways of selecting pairs (two boys) of boys from a group of

four is given using combination formula as follows;

$_nC_r = \dfrac{n!}{r! \cdot (n - r)!}$

Which gives;

$_4C_2 = \dfrac{4!}{2! \cdot (4 - 2)!} = 6$

₄C₂ = 6

Therefore, the weights of the pairs are;

a + b = 110

a + c = 112

a + d = 113

b + c = 118

b + d = 121

c + d = Lost weight

c + d = Lost weight

a + b + c + d = 230 (given)

Therefore;

The lost weight, c + d = 230 - (a + b)

Which gives;

c + d = 230 - (110) = 120

The lost weight = c + d = 120

The lost weight = 120 kg

Learn more here:

brainly.com/question/12974932

6 0

3 years ago

Read 2 more answers

What number is 8.2% of 50

Svet_ta [14]

Your answer would be 4.1

4 0

3 years ago

Read 2 more answers

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) 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.

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

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

5 0

4 years ago