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galben [10]
3 years ago
5

A book exhibition was held for four days in a schoolthe number of tickets sold at the counter on the first secend third and four

th final day was respectivally 1094 1812 2050 and 2751 find the total number of tickets sold on all the four days
Mathematics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

7707 tickets

Step-by-step explanation:

From the question we know that:

the number of tickets sold at the counter on

First day = 1094 tickets

Second day = 1812 tickets

Third day = 2050 tickets

Fourth day = 2751 tickets

The total number of tickets sold on all the four days is calculated as:

(1094 + 1812 + 2050 + 2751)tickets

= 7707 tickets

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What postulate or theorem can be used to prove that these two triangles are congruent?
Diano4ka-milaya [45]

Answer:

ASA

Step-by-step explanation:

in the place that I have put some marks on it, they have the same angle; so then they will also have the same third angle; and for the rest you should use ASA.

8 0
2 years ago
Harmonic mean if a number h such that (h-a)/(b-h)=a/b Prove h is H(a,b) iff satisfies either relation a. (1/a)-(1/h) = (1/h) - (
Fudgin [204]
A.

\displaystyle\frac1a-\frac1h=\frac1h-\frac1b
\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}
\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}
\implies\displaystyle\frac{h-a}{b-h}=\frac ab

b.

\displaystyle h=\frac{2ab}{a+b}
\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}
\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}
\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}
\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}
\displaystyle\implies\frac{h-a}{b-h}=\frac ab

The other direction can be proved by following the manipulations in the reverse order.
7 0
3 years ago
Needdd hellpppppssssssss
Ratling [72]

Answer:

Choice number one:

\displaystyle \frac{5}{10}\cdot \frac{4}{9}.

Step-by-step explanation:

  • Let A be the event that the number on the first card is even.
  • Let B be the event that the number on the second card is even.

The question is asking for the possibility that event A and B happen at the same time. However, whether A occurs or not will influence the probability of B. In other words, A and B are not independent. The probability that both A and B occur needs to be found as the product of

  • the probability that event A occurs, and
  • the probability that event B occurs given that event A occurs.

5 out of the ten numbers are even. The probability that event A occurs is:

\displaystyle P(A) = \frac{5}{10}.

In case A occurs, there will only be four cards with even numbers out of the nine cards that are still in the bag. The conditional probability of getting a second card with an even number on it, given that the first card is even, will be:

\displaystyle P(B|A) = \frac{4}{9}.

The probability that both A and B occurs will be:

\displaystyle P(A \cap B) = P(B\cap A) =  P(A) \cdot P(B|A) = \frac{5}{10}\cdot \frac{4}{9}.

6 0
2 years ago
Bob walked 20 miles in 4 hours. Kelly walked 14 miles in 3 hours. Are these rates proportional
zaharov [31]

Answer:

No

Step-by-step explanation:

20/4=5

14/3=4.67

Since they're not the same, they're not proportional.

7 0
3 years ago
Read 2 more answers
Someone please help me this is due today pleasee
kolbaska11 [484]

Answer:

A and D

Step-by-step explanation:

yes

4 0
3 years ago
Read 2 more answers
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