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3 years ago

Convert 715% to a decimal number

Mathematics

2 answers:

Arlecino [84]3 years ago

6 0

$715\%=7.15$

Aleksandr [31]3 years ago

4 0

715%=7 3/20=7.15
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Use rounding numbers to estimate the sum of 171 + 727

Umnica [9.8K]

171 rounds to 200, and 727 rounds to 700. 700+200= 900.

The answer is 900.

5 0

3 years ago

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Let Y be a random variable with a density function given by

Neporo4naja [7]

From the given density function we find the distribution function,

$F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y$

(a)

$F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)$

$\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1$

$\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}$

(b)

$F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y$

$\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2$

$\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}$

(c)

$F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})$

$\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3$

$\implies f_{U_3}(u_3}=\begin{cases}\frac32\sqrt u&\text{for }0\le u\le1\\0&\text{otherwise}\end{cases}$

5 0

3 years ago

For triangle ABC, angle A=110, angle B=43, and side b=15. Find side c.

vredina [299]

Answer: 10

Step-by-step explanation:

First, find m∠C by using the Triangle Sum Theorem:

∠A + ∠B + ∠C = 180°

110 + 43 + ∠C = 180°

153 + ∠C = 180°

∠C = 27°

Next, Use the Law of Sines to solve for c:

$\dfrac{b}{sin\ B}= \dfrac{c}{sin\ C}$

$\dfrac{15}{sin\ 43}= \dfrac{c}{sin\ 27}$

$\dfrac{15\ sin\ 27}{sin\ 43}= c$

$10 = c$

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3 years ago

ASAP PLZZZZZZ NOW ! answer all questions

REY [17]

Answer:

68km

22miles

240km

Step-by-step explanation:

6 0

3 years ago

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There are 15 candidates for 4 job positions.

Veronika [31]

Answer:

32760 ways

Step-by-step explanation:

Given

Number of Candidates = 15

Job Positions = 4

Required:

Number of outcomes

This question represent selection; i.e. selecting candidates for job positions;

This question can be solved in any of the following two ways

Method 1.

The first candidate can be chosen from any of the 15 candidates

The second candidate can be chosen from any of the remaining 14 candidates

The third candidate can be chosen from any of the remaining 13 candidates

The fourth candidate can be chosen from any of the remaining 12 candidates

Total Possible Selection = 15 * 14 * 13 * 12

Total Possible Selection = 32760 ways

Method 2:

This can be solved using permutation method which goes thus;

$nPr = \frac{n!}{(n-r)!}$

Where n = 15 and r = 4

So;

$nPr = \frac{n!}{(n-r)!}$ becomes

$15P4 = \frac{15!}{(15-4)!}$

$15P4 = \frac{15!}{11!}$

$15P4 = \frac{15*14*13*12*11!}{11!}$

$15P4 = 15*14*13*12$

$15C4 = 32760$

Hence, there are 32760 ways

8 0

3 years ago