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2 years ago

12

(9.05x 104)*(3.2x10-3) =

1 answer:

Andru [333]2 years ago

7 0

Step-by-step explanation:

$\tt{}(9.05 \times 104) \times (3.2 \times 10 - 3)$

$\tt{}941.2 \times 29$

$\tt{}27294.8$

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In a given binomial distribution, there are 625 trials, and the probability of success is p =0.29. What is the variance?

Shkiper50 [21]

Answer:

128.69.

Step-by-step explanation:

The variance is np(1 - p)

= 625*(0.28) * 0.71

= 128.69.

3 0

3 years ago

Which set of measurements could be the side lengths of a triangle

inessss [21]

Answer:

Hi,

Step-by-step explanation:

(13,7,4) may not be a triangle since 13 is not little than 7+4=11
(7,10,8,) may be a triangle (7 < 10+8 , 10 < 7+8, 8 < 10+7)

(9,4,4) no since 9 is not < 4+4

(2,8,12) no since 12 is not < 2+8

The only answer is (7,10,8)

7 0

1 year ago

Divide 3520 in the ratio 11:12:17

Alenkasestr [34]

Answer:

(11+12+17)x = 40x.

Step-by-step explanation:

Rs 3520 is divided in the Ratio 11:12:17 as Rs 968, Rs 1056, Rs 1496. According to the question, we have to divide Rs 3520 in the ratio 11:12:17. Let the common ratio be x. So, total share = (11+12+17)x = 40x.

4 0

3 years ago

Read 2 more answers

Express the number 220 as the sum of four numbers that form a geometric progression such that the third term is greater than the

siniylev [52]

Answer:

Step-by-step explanation:

The first four terms of geometric series is:

$a,ar,ar^2,ar^3$

Since, we have given information that the third number is greater than 44 that means:

$ar^2=a+44$

Above equation can be rewritten as:

$a(r^2-1)=44$

Now, using:

$a^2-b^2=(a+b)(a-b)$

Here, a=r,b=1

$a(r+1)(r-1)=44$ (1)

The sum of first four terms is:

$a+ar+ar^2+ar^3=220$

$a(1+r+r^2+r^3)=220$

$a(1(1+r)+r^2(1+r))=220$

$a((1+r)(1+r^2))=220$ (2)

Divide equation (2) by (1) we get:

$\frac{r^2+1}{r-1}=\frac{220}{44}$

$r^2+1=5r-5$

$\Rightarrow r^2-5r+6=0$

$\Rightarrow r^2-3r-2r+6=0$

$\Rightarrow r(r-3)-2(r-3)=0$

$\Rightarrow (r-2)(r-3)=0$

$\Rightarrow r=2,3$

CASE1: When r=2 in $ar^2=a+44$

$a(4)=a+44$

$3a=44$

$a=\frac{44}{3}$

CASE2:When r=3 in $ar^2=a+44$

$a(3)^2=a+44$

$\Rightarrow 9a=a+44$

$\Rightarrow 8a=44$

$\Rightarrow a=\frac{44}{8}=\frac{11}{2}$

The series becomes:

From CASE1: $\frac{44}{3},\frac{44\cdot 2}{3},\frac{44\cdot 2^2}{3},\frac{44\cdot 2^3}{3}$

$\Rightarrow \frac{44}{3},\frac{88}{3},\frac{176}{3},\frac{352}{3}$

From CASE2: $\frac{11}{2},\frac{11\cdot 3}{2},\frac{11\cdot 3^2}{2},\frac{11\cdot 3^3}{2}$

$\Rightarrow \frac{11}{2},\frac{33}{2},\frac{99}{2},\frac{297}{2}$

6 0

3 years ago

Read 2 more answers

Can somebody help me do this plz?

sdas [7]

Here’s my answer, hope this helps!

4 0

3 years ago

Read 2 more answers