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Otrada [13]
3 years ago
14

Need help please, thanks!

Mathematics
1 answer:
MakcuM [25]3 years ago
4 0
Given:
1 inch : 18 kilometers

Let us use the ratio and proportion method:

a:b = c:d where ad = bc

0.8 inches is <u>14.4 kilometers</u>

1 : 18 = 0.8 : x
x = 0.8 * 18
x = 14.4 km

1.4 inches is <u>25.2 kilometers</u>

1: 18 = 1.4 : x
x = 18 * 1.4
x = 25.2 km

2.1 inches is <u>37.8 kilometers</u>

1 : 18 = 2.1 : x
x = 18 * 2.1
x = 37.8 km

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According to a recent study, the carapace length for adult males of a certain species of tarantula are normally distributed with
seraphim [82]
Idk tbh sorry bye ok hbye
3 0
3 years ago
The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
3 years ago
suppose x and y vary inversely and y=10 when x=4. write a function that models the variation then find y when x =90
ankoles [38]
X = k/y
4 = k/10
k = 4 x 10 = 40
Thus, x = 40 / y

90 = 40 / y
y = 40/90 = 4/9
4 0
3 years ago
Need help with this math
Oksi-84 [34.3K]
Answer:
B. 6 + √26
Step-by-step explanation:
1. Make a right triangle using the line that goes from office to supermarket.
2. Use pythagorean thereom to find the distance of the longest side.
3. FInd the distance of the line segment going from supermarket to Marias House.
4. Add them together to get 6 + √26.
6 0
3 years ago
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dino has 14 coins and 2 one dollar bills which describes the relationships between the coins and the dollar bills
kkurt [141]

Answer:

they are both money

Step-by-step explanation:

that's why they relate, they're both currency

6 0
3 years ago
Read 2 more answers
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