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poizon [28]
4 years ago
11

How many odd numbers with 4 different digits, can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8? (No repetition is allowed)

Mathematics
1 answer:
olga55 [171]4 years ago
6 0

Answer:

840 ( D )

Step-by-step explanation:

GIVEN DIGITS : 1,2,3,4,5,6,7,8  

Number of odd numbers = 4

Number of even numbers = 4

therefore the number of odd numbers with 4 different digits can be formed by the same way the number of even numbers ( without repetition )

Hence the number of ways odd numbers with 4 different digits = Total number of ways of forming 4 digit numbers / 2

8*7*6*5 = 1680 / 2 =  840 ways

You might be interested in
determina la medida de cada uno de los angulos del triángulo ABC si se conoce que (3x+1)° (3x+4)° y (2x-1)°​
Triss [41]

Answer:

∠A=67°

∠B=70°

∠C=43°

Step-by-step explanation:

<u><em>The correct question in English is</em></u>

Determines the measurement of each of the angles of the triangle ABC if it is known that A=(3x+1)°,B=(3x+4)° and C=(2x-1)°.

we know that

The measure of the interior angles in a triangle must be equal to 180 degrees

so

∠A+∠B+∠C=180°

substitute the given values

(3x+1)\°+(3x+4)\°+(2x-1)\°=180\°

solve for x

(8x+4)\°=180\°

8x=180-4

x=22

Find the measure of each angle

∠A=(3(22)+1)=67°

∠B=(3(22)+4)=70°

∠C=(2(22)-1)=43°

6 0
3 years ago
What is the answer to –(d+1)&gt;3
aleksandrvk [35]

Answer:

−(d+1)>3

Step 1: Simplify both sides of the inequality.

−d−1>3

Step 2: Add 1 to both sides.

−d−1+1>3+1

−d>4

Step 3: Divide both sides by -1.

\frac{-d}{-1} >  \frac{4}{-1}

d < -4

5 0
3 years ago
Venla is 5 years older than her cousin Kora. Write an equation for the age of Venla, v, when Kora is k years old.
Talja [164]
K + 5 =V
Kiena + 5 years = Venus’s age
5 0
4 years ago
Read 2 more answers
Alice has a total of 12 dimes and nickels. She has 2 more nickels than dimes. Which equation represents the given problem situat
morpeh [17]

Answer:

your answer should be 3

5 0
3 years ago
At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37
monitta

Answer:

Half-life of the goo is 49.5 minutes

G(t)= 300e^{-0.014t}

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let M_0\,,\,M_f denotes initial and final mass.

M_0=300\,\,grams\,,\,M_f=37.5\,\,grams

According to exponential decay,

\ln \left ( \frac{M_f}{M_0} \right )=-kt

Here, t denotes time and k denotes decay constant.

\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014

So, half-life of the goo in minutes is calculated as follows:

\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes

Half-life of the goo is 49.5 minutes

\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}

So,

G(t)= M_f=M_0e^{-kt}

Put M_0=300\,\,grams\,,\,k=0.014

G(t)= 300e^{-0.014t}

Put t = 32 minutes

G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams

7 0
3 years ago
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