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2 years ago

Could you tell me the answer please it is not multiple-choice

Mathematics

1 answer:

skad [1K]2 years ago

8 0

Answer: a) 300000 b) 200 c) 20000

Step-by-step explanation:

Round so that there is only 1 sig. fig. in the answer

a) 300000 (rounded up to 300,000 not 280,000 as there would be 2 sig figs, 2 and 8)

b) 200 (rounded down)

c) 20000 (rounded up to 20,000 not 18,000 as there would be 2 sig figs, 1 and 8)

The trailing zeroes do not count as sig. figs. so the above answers all have only one sig. fig.

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a map is drawn using a scale of 1cm/80mi two cities are 280 miles apart how far apart are the cities on the map

anastassius [24]

3 1/2 cm or 3.5 cm. Because you should divide 280 by 80 which equals 3.5. turn 3.5 to centimeters. Thats your answer.

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2 years ago

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17. What are the three angle measures in the triangle shown below?

lianna [129]

Answer:

Option (1)

Step-by-step explanation:

By using triangle sum theorem in the given triangle,

(5x - 7)° + (11x - 4)° + (3x + 1)° = 180°

(5x + 11x + 3x) - (7 + 4 - 1) = 180

19x - 10 = 180

19x = 190

x = 10

Therefore, (3x + 1)° = 30 + 1

= 30°

(11x - 4)° = 110 - 4

= 106°

(5x - 7)° = 43°

Therefore, Option (1) will be the correct option.

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2 years ago

IF I HAVE A LAPTOP COMPUTER THAT MEASURES 30.8×6.6×39.8 CENTIMETERS...WHAT SIZE CASE DOES IT REQUIRE

soldi70 [24.7K]

Answer:

8090.54

Step-by-step explanation:

30.8*6.6*39.8

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2 years ago

Add the following complex numbers:<br> (6 - 21) + (11 +61)

umka2103 [35]

Answer:

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Step-by-step explanation:

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2 years ago

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

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2 years ago