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3 years ago

Convert 78% to a decimal

Mathematics

2 answers:

pychu [463]3 years ago

8 0

The correct answer is 0.78. Hope this helps.

aleksandrvk [35]3 years ago

7 0

If you divide a percent by 100 it will give you the decimal so 78% would be 0.78

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www.g A bag contains 3 white counters, 10 black counters, and 4 green counters. What is the probability of drawing (a) a white c

erastovalidia [21]

Answer:

41.18% probability of drawing a white counter or a green counter

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question:

There are 3+10+4 = 17 counters.

Of those, 3+4 = 7 are white or green

7/17 = 0.4118

41.18% probability of drawing a white counter or a green counter

4 0

3 years ago

73.50 divided by 21. Ishsdbdh

valina [46]

3.5 is the answer!!!

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2 years ago

Read 2 more answers

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

3 years ago

I’ll CashApp someone 5$ if they can kindly answer this, I’ve been stuck on this and it’s due tonight, I will even mark you Brain

lisov135 [29]

Answer: Part A is 2 and 6 Part B is 2

Step-by-step explanation:

Part A: Here is the explanation. So, you started at with the expression 3x^2+8x+4 and when you're are factoring, you have 3x^2+px+pq+4. You can substitute the p and q for 6 and 2. What they did is they replaced 8x with px+qx. To get 8x, p needs to be 6 and q needs to be 2, or the other way around. TIP: The numbers just have to add up to 8 on this one. It doesn't have to be 6 and 2.

Part B: Here is what I got so far... 3x(x+r) is 3x^2+3xr. Also, s(x+r) is sx+sr. The equation becomes, 3x^2+3xr+sx+sr. R can be 2 and s can be 2. Here is my reasoning: The original expression was 3x^2+8x+4. We already have the 3x^2, so now we need to find what the others are by determining what r and s equal. R and s can both be 2 to make four. 2x2 is 4. Let's see if it can make 8. 3xr becomes 6x and sx becomes 2x. 6x+2x is 8x.

4 0

3 years ago

The system of equation 3x-6y=30 and 2x-4y=3 is

Anna11 [10]

No solution as they are parallel lines. so, inconsistent

6 0

3 years ago