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nadezda [96]
3 years ago
14

Is (0,0) a solution to the system

Mathematics
2 answers:
lisov135 [29]3 years ago
7 0

Answer:

Yes it could be a solution

Step-by-step explanation:

constants become 0 as variables.  0 likewise other than constants become 0

constants are 0.

Maru [420]3 years ago
3 0

Answer:

yes (0,0) can be solution to system

when constant becomes 0 as variables becomes 0 likewise other terms except constant becomes 0 so to satisfy equation ,constant should be 0

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the football team has 4 freshman, 7 sophomores, 10 juniors, and 15 seniors. two players will be chosen at random to participate
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This answer would have to be snakskeewhsof ndxjssoaanshew
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3 years ago
Find the product of (5.2*10^-6) and (8*10^3)
Maru [420]

Answer:

8000.0000052

Step-by-step explanation:

(5.2*10^-6)=0.0000052

(8*10^3)=8000

3 0
3 years ago
WILL MARK BRAINIEST Bob cannot decide which of two washing machines to buy. The selling price of each is ​$495. The first is mar
sashaice [31]

Answer:

See explanation

Step-by-step explanation:

495*.3= 148.5

<u>346.50</u>

495-99(20%)

396-10%

<u>356.4</u>

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3 0
3 years ago
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fredd [130]

Answer:

A

Step-by-step explanation:

28 + 8 is 26

6 0
3 years ago
A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identi
Tcecarenko [31]

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

4 0
3 years ago
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