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3 years ago

Help me ans should be well explained

Mathematics

2 answers:

Katen [24]3 years ago

8 0

Finding diagonal sum

$\\ \sf\longmapsto 0.7+1.0+1.3$

$\\ \sf\longmapsto 0.7+2.3$

$\\ \sf\longmapsto 3.0$

Observing the square

x=0.9
y=0.6

$\\ \sf\longmapsto xy=0.9(0.6)$

$\\ \sf\longmapsto xy=0.54$

Ierofanga [76]3 years ago

4 0

Answer:

=====> A

⟼0.7+1.0+1.3

⟼0.7+2.3

⟼3.0

======> B

x=0.9

y=0.6

⟼xy=0.9(0.6)

⟼xy=0.54

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3 years ago

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Write an equation of the line with a slope of 3 and y-intercept of -8

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Answer:

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Step-by-step explanation:

3 0

3 years ago

Q4.

goblinko [34]

The coding of the statistic is used to make it easier to work with the large sunshine data set

The mean of the sunshine is 3.05 $\overline 6$

The standard deviation is approximately 18.184

Reason:

The given parameters are;

The sample size, n = 3.

∑x = 947

Sample corrected sum of squares, Sₓₓ = 33,065.37

The mean and standard deviation = Required

Solution:

$Mean, \ \overline x = \dfrac{\sum x_i}{n}$

The mean of the daily total sunshine is therefore;

$Mean, \ \overline x = \dfrac{947}{30} \approx 31.5 \overline 6$

$s = \dfrac{x}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{Ex}{10 } - \dfrac{1}{10}$

$E(s) = \dfrac{31.5 \overline 6}{10 } - \dfrac{1}{10} = 3.05 \overline 6$

The mean ≈ 3.05 $\overline 6$

Alternatively

, $The \ mean \ of \ the \ daily \ total \ sunshine, \, s = \dfrac{31.5 \overline 6 - 1}{10 } = 3.05\overline 6$

The mean of the daily total sunshine, $\overline s$ ≈3.05 $\overline 6$

$Var(s) = Var \left(\dfrac{x}{10 } - \dfrac{1}{10} \right)$

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times Var \left(x \right)$

Therefore;

$Var(s) = \left(\dfrac{1}{10}\right)^2 \times 33,065.37 = 330,6537$

Therefore;

$s = \sqrt{330.6537} \approx 18.184$

The standard deviation, $s_s$ ≈ 18.184

Learn more about coding of statistic data here:

brainly.com/question/14837870

3 0

2 years ago

Fifty pro-football rookies were rated on a scale of 1 to 5, based on performance at a training camp as well as on past performan

son4ous [18]

Answer:

(a-1) 22 rookies receiving a rating of 4 or better.

(a-2) 14 rookies received a rating of 2 or worse.

(b-1) Constructed below in explanation.

(b-2) 8% of total rookies received a rating of 5.

Step-by-step explanation:

We are provided the rating of Fifty pro-football rookies on a scale of 1 to 5 based on performance at a training camp as well as on past performance.

The frequency distribution constructed is given below:

Rating Frequency

1 4

2 10 where ranking of 1 indicate a poor prospect whereas

3 14 ranking of 5 indicate an excellent prospect.

4 18

5 4

(a-1) Rookies receiving a rating of 4 or better = Rating of 4 + Rating of 5

So, by seeing the frequency distribution 18 rookies received a rating of

4 and 4 rookies received a rating of 5.

Hence, Rookies receiving a rating of 4 or better = 18 + 4 = 22 rookies.

(a-2) Number of rookies received a rating of 2 or worse = Rating of 2 +

Rating of 1

So, by seeing the frequency distribution 10 rookies received a rating of

2 and 4 rookies received a rating of 1.

Hence, Rookies receiving a rating of 2 or worse = 10 + 4 = 14 rookies.

(b-1) Relative Frequency is calculated as = Each frequency value /

Total Frequency

Rating Frequency(f) Relative Frequency

1 4 4 / 50 = 0.08

2 10 10 / 50 = 0.2

3 14 14 / 50 = 0.28

4 18 18 / 50 = 0.36

5 4 4 / 50 = 0.08

$\sum f$ = 50

Hence, this is the required relative frequency distribution.

(b-2) To calculate what percent received a rating of 5 is given by equation :

x% of 50 = 4 {Here 4 because 4 rookies received rating of 5}

x = $\frac{4*100}{50}$ = 8% .

Therefore, 8% of total rookies received a rating of 5.

5 0

4 years ago

Can somebody help please!????

stiv31 [10]

A and b proves that they are parallel

4 0

3 years ago

Help me ans should be well explained ​

Help me ans should be well explained