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kenny6666 [7]
2 years ago
7

Which of the following is true about a parallelogram?

Mathematics
2 answers:
andriy [413]2 years ago
5 0
The diagonals bisect each other
Lerok [7]2 years ago
5 0

Answer:

A paralellogram has 2 pair of equal sides

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0<br> (4<br> Зу = 4х + 15 аnd 9x + 12y = 12
lys-0071 [83]

Step-by-step explanation:

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6 0
3 years ago
On Friday, Leslie sold 9 3/8 pitchers of lemonade from her lemonade stand. On Saturday, she sold 3 times as much lemonade as on
shtirl [24]

Answer:

28 1/8

Step-by-step explanation:

8 0
3 years ago
There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probabilit
Kaylis [27]

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a \frac{3}{11} chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is \frac{8}{11}, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is\frac{2}{11}, and the probability of picking the second red ball is \frac{1}{10} and the probability of picking the green ball is\frac{1}{9}. We want to multiply thm again, as per the multiplication rule like the last problem.

\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}

5 0
3 years ago
Some transportation experts claim that it is the variability of speeds, rather than the level of speeds, that is a critical fact
scZoUnD [109]

Answer:

Explained below.

Step-by-step explanation:

The claim made by an expert is that driving conditions are dangerous if the variance of speeds exceeds 75 (mph)².

(1)

The hypothesis for both the test can be defined as:

<em>H</em>₀: The variance of speeds does not exceeds 75 (mph)², i.e. <em>σ</em>² ≤ 75.

<em>Hₐ</em>: The variance of speeds exceeds 75 (mph)², i.e. <em>σ</em>² > 75.

(2)

A Chi-square test will be used to perform the test.

The significance level of the test is, <em>α</em> = 0.05.

The degrees of freedom of the test is,

df = n - 1 = 55 - 1 = 54

Compute the critical value as follows:

\chi^{2}_{\alpha, (n-1)}=\chi^{2}_{0.05, 54}=72.153

Decision rule:

If the test statistic value is more than the critical value then the null hypothesis will be rejected and vice-versa.

(3)

Compute the test statistic as follows:

\chi^{2}=\frac{(n-1)\times s^{2}}{\sigma^{2}}

    =\frac{(55-1)\times 94.7}{75}\\\\=68.184

The test statistic value is, 68.184.

Decision:

cal.\chi^{2}=68.184

The null hypothesis will not be rejected at 5% level of significance.

Conclusion:

The variance of speeds does not exceeds 75 (mph)². Thus, concluding that driving conditions are not dangerous on this highway.

7 0
3 years ago
Fifty eight students were surveyed about how frequently they text during lunch. The results are summarized below. Which statemen
umka2103 [35]

Answer:The median number is 9 , Q3 is 11, data is skewed left, lowest 25%

Step-by-step explanation:

4 0
3 years ago
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