Please help me! I will give brainly!!!

5. The superintendent of the local school district claims that the children in her district are brighter, on average, than the g

anygoal [31]

Answer:

We conclude that children in district are brighter, on average, than the general population.

Step-by-step explanation:

We are given the following data set:

105, 109, 115, 112, 124, 115, 103, 110, 125, 99

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.

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

$Mean =\displaystyle\frac{1117}{10} = 111.7$

Sum of squares of differences = 642.1

$S.D = \sqrt{\frac{642.1}{49}} = 8.44$

We are given the following in the question:

Population mean, μ = 106

Sample mean, $\bar{x}$ = 111.7

Sample size, n = 10

Alpha, α = 0.05

Sample standard deviation, s = 8.44

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 106\\H_A: \mu > 106$

We use one-tailed(right) t test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$t_{stat} = \displaystyle\frac{111.7 - 106}{\frac{8.44}{\sqrt{10}} } = 2.135$

Now,

$t_{critical} \text{ at 0.05 level of significance, 9 degree of freedom } = 1.833$

Since,

$t_{stat} > t_{critical}$

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

We conclude that children in district are brighter, on average, than the general population.

4 0

3 years ago

Given the point A(-3,-2) and B(6,1), select two possible locations for point P so that P partitions AB in the ratio 2:1

natta225 [31]

Answer: (0,3) & (0,-1)

Step-by-step explanation:

7 0

4 years ago

The accompanying data contains the depth (in kilometers) and magnitude, measured using the Richter Scale, of all earthquakes

Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ = 34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth ${}$ Magnitude

0.76 ${}$ 0.84

4.93 ${}$ 0.47

8.16 ${}$ 0.35

33.58 ${}$ 1.32

21.2 ${}$ 1.61

35.03 ${}$ 4.57

10.05 ${}$ 5.52

47.91 ${}$ 1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}$

Where;

$x_i$ = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

$\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025 \right )^{2} }{8}} \approx 15.92 \ km$

Q₁ = The first quartile = The (n + 1)/4th = term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th = term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

4 0

4 years ago

Help pls :’( ASAPP!!!<br> “Complete the proof”

nata0808 [166]

1) $\overline{AB} \cong \overline{CD}$ , $\overline{AD} \cong \overline{CB}$ , $\overline{AX} \perp \overline{BD}$ , $\overline{CY} \perp\overline{BD}$ (given)

2) $\overline{BD} \cong \overline{BD}$ (reflexive property)

3) $\triangle ABD \cong \triangle ACDB$ (SSS)

4) $\angle ADB \cong \angle CBY$ (CPCTC)

5) $\angle CYB$ and $\angle AXD$ are right angles (perpendicular lines form right angles)

6) $\triangle CYB$ and $\triangle AXD$ are right triangles (a triangle with a right angle is a right triangle)

7) $\triangle AXD \cong \triangle CYB$ (HA)

8) $\overline{AX} \cong \overline{CY}$ (CPCTC)

6 0

2 years ago

What are the characteristics of x^2-x+1

SSSSS [86.1K]

It is a quadratic equation, if it is equal to zero. It is not factor able.

7 0

4 years ago