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skad [1K]
2 years ago
11

If you want 10 points answer my question Plz

Mathematics
1 answer:
lara31 [8.8K]2 years ago
8 0
The mean salary would become $11.50.
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148^2

Step-by-step explanation:

2(lb+bh+lh) = 2(4*5+5*6+4*6)=148 in^2

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2 years ago
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2.74 as a mix number
Inga [223]

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2 74/100

Step-by-step explanation:

6 0
3 years ago
How do i find a formula for "1+3+5+7+......+(2n-1)
nata0808 [166]

Answer:

Step-by-step explanation:

common difference d=3-1=2

first term a=1

an=a+(n-1)d

2n-1=1+(l-1)2

2n-1=1+2l-2

2n-1=2l-1

l=n

(i used l for number of terms)

number of terms=n

S_{n}=\frac{n}{2} (first ~term+last~term)\\=\frac{n}{2} (1+2n-1)\\=n^2

7 0
2 years ago
Which is the approximate measure of this angle A. 30 b. 90 c 120 d 180
statuscvo [17]

Answer:

Answer is A. 30 degree

as we know that two prependicular lines have 90 degree angle between them

and similarly a horizontal line has angle zero degree.

the angle of pic shown is approximately 30 degree as it lies between prependicular and horizontal nature of lines

but as we can see that it is close to horizontal nature of the line that's why its approximate angle is 30 degree.

4 0
3 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} 

2 =   e^{ \alpha t} 

 \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
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