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marusya05 [52]
3 years ago
13

Find the missing measure.

Mathematics
1 answer:
eduard3 years ago
4 0

Answer:

59

Step-by-step explanation:

(360-92-150)/2

= 118/2

= 59

Answered by GAUTHMATH

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Geometry/// find the surface area of the square pyramid if the height of each triangular face is 20 m and the length of each sid
STALIN [3.7K]

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1202.38

Step-by-step explanation:

i looked how how do it and a calculator popped up so i put the numbers in that's what i got

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3/5x + 35°<br>Find the value of x.<br>135<br>145<br>quizzes ​
lapo4ka [179]

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⅕x (3x + 175)

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3 years ago
Choose the numbers that are terminating decimals. Select all that apply.
horsena [70]

Answer: A, C, and D

Step-by-step explanation:

          A terminating decimal is a decimal that has an end. In other words, \frac{1}{4}  =0.25 is one, but \frac{1}{3} =0.3333... is not.

✓ A. 0.032

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3 0
2 years ago
A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, ev
Olin [163]

Answer:

The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Step-by-step explanation:

Given : A certain class has 20 students, and meets on Mondays and Wednesdays in a classroom with exactly 20 seats. In a certain week, everyone in the class attend both days. On both days, the students choose their seats completely randomly (with one student per seat).

To find : The probability that no one sits in the same seat on both days of that week ?

Solution :

Let A_i be the i-th student sits on seat which he has been sitting on Monday.

According to question,

We have to calculate P(\cap^{20}_{i=1}A_i^c)

Applying inclusion exclusion formula,

P(\cap^{20}_{i=1}A_i^c)=1-P(\cap^{4}_{i=1}A_i)

P(\cap^{20}_{i=1}A_i^c)=1-P(A_1)+...+P(A_{20})-P(A_1\cap A_2)+...+P(A_{19}\cap A_{20})+P(A_1\cap A_2\cap A_3)+...+P(A_{18}\cap A_{19}\cap A_{20})....-P(A_1\cap A_2...\cap A_{20})

Using symmetry,

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}P(A_1\cap ...\cap A_k)

P(\cap^{20}_{i=1}A_i^c)=1-\sum^{20}_{k=1}(-1)^{k+1}\binom{20}{k}\frac{(20-k)!}{20!}

P(\cap^{20}_{i=1}A_i^c)=1+\sum^{20}_{k=1}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\sum^{20}_{k=0}(-1)^{k}\frac{1}{k!}

P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

Therefore, The probability that no one sits in the same seat on both days of that week is given by, P(\cap^{20}_{i=1}A_i^c)=\frac{1}{e}

8 0
3 years ago
Proportional to 40⁄8?
murzikaleks [220]

Answer:

5

Step-by-step explanation:

DIVIDE EACH SIDE BY 8 OR DO 40 DIVIDED BY 8

6 0
3 years ago
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