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Grace [21]
3 years ago
8

What is 5/2 times x =1

Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer:

2/5

Step-by-step explanation:

5/2  its so simple 5 x 2 is 10 /2 x 5 is 10 and 10/10 is 1


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Y=145, I’m not sure about X

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Which equation is represented by the phrase “one-fourth of a number, increased by eight equals sixteen”?
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Answer:

One-fourth n + 8 = 16

Step-by-step explanation:

basically 1/4n +8=16

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3 years ago
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Suppose the number of radios in a household has a binomial distribution with parameters n=11 and p=40%. Find the probability of
Ivenika [448]

Answer:

Step-by-step explanation:

The formula for binomial distribution is expressed as

P(x=r) = nCr × q^(n-r) × p^r

From the information given,

n = 11

p = 40% = 40/100 = 0.4

q = 1 - 0.4 = 0.6

x represent the number of radios

a) P( x = 1) or P(x = 9)

P(x = 1) = 11C1 × 0.6^(11-1) × 0.4^1

P(x = 1) = 0.027

P(x = 9) = 11C9 × 0.6^(11-9) × 0.4^9

P(x = 9) = 0.0052

P( x = 1) or P(x = 9) = 0.027 + 0.0052 = 0.0322

b) P(x lesser than or equal to 7) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)

P(x = 0) = 11C0 × 0.6^(11-0) × 0.4^0 = 0.004

P(x = 1) = 0.027

P(x = 2) = 11C2 × 0.6^(11-2) × 0.4^2 = 0.089

P(x = 3) = 11C3 × 0.6^(11-3) × 0.4^3 = 0.177

P(x = 4) = 11C4 × 0.6^(11-4) × 0.4^4 = 0.24

P(x = 5) = 11C5 × 0.6^(11-5) × 0.4^5 = 0.22

P(x = 6) = 11C6 × 0.6^(11-6) × 0.4^6 = 0.15

P(x = 7) = 11C7 × 0.6^(11-7) × 0.4^7 = 0.15

P(x lesser than or equal to 7) = 0.004 + 0.027 + 0.089 + 0.177 + 0.24 + 0.22 + 0.15 + 0.07 = 0.977

c) P(x greater than or equal to 5) = 1 - P(x lesser than or equal to 4) = 1 - (0.004 + 0.027 + 0.089 + 0.177 + 0.24) = 1 - 0.537 = 0.463

d) P(x lesser than 9) = P(x lesser than or equal to 7) + P(x = 8)

P(x = 8) = 11C8 × 0.6^(11-8) × 0.4^8 = 0.02

P(x lesser than 9) = 0.977 + 0.02 = 0.997

e. P(x greater than 7) = 1 - P(x lesser than or equal to 7) = 1 - 0.977 = 0.023

5 0
3 years ago
In a function if the vaule of a changes in response to the vaule of b then b is the
d1i1m1o1n [39]
<span>b is the independent variable ok</span>
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3 years ago
Niklas takes a dose of 252525 micrograms (\text{mcg})(mcg)left parenthesis, m, c, g, right parenthesis of a certain supplement e
11Alexandr11 [23.1K]

Niklas takes a dose of 25 micrograms of a certain supplement each day. The supplement has a half life of 4 hours, meaning that 1/64 of the supplement remains in the body after each day. How much of the supplement is in Niklas's body immediately after the 12th dose? Round your final answer to the nearest hundredth.

Answer:

The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth

Step-by-step explanation:

Half life is the time required for an element to decay into half of its initial size.

Given that :

The supplement has a half life of 4 hours, this implies that  it decay to half of its size every 4 hours.

∴ there are 6 stages of division in a day.

i.e

\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2} =\dfrac{1}{64}

The amount of the supplement in Niklas body after the first dose (first day) can be calculated as:

=  \dfrac{1}{64} × 25

= 0.390625 micrograms

It is said that he used the supplement daily for 12 days (12th dose),

As such ; we can estimate the amount of the supplement that is in his body immediately after the 12th dose; which is calculated as:

amount in his body per day × number of period for complete decay

= 0.390625 × 11

= 4.296875

≅4.30 micrograms

= 430 micrograms to the nearest hundreth

The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth

3 0
2 years ago
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