Ask question

Ask question

All categories

Ira Lisetskai [31]

3 years ago

12

Stephen uses \dfrac{2}{25}

1 answer:

zepelin [54]3 years ago

6 0

Answer:

$13\dfrac{3}{4}$ servings$

Step-by-step explanation:

Stephen uses $\dfrac{2}{25}$ kilograms of tofu in each serving of his dish.

He has $1\dfrac{1}{10}$ kilograms of tofu.

We want to determine how many servings Stephen can make.

To do this, we divide the total weight of tofu by the weight per serving.

$\dfrac{\text{Total Weight(in kg)}}{\text{Weight(kg) per serving}} =\dfrac{1\dfrac{1}{10}}{ \dfrac{2}{25}}\\ = \dfrac{11}{10}\div \dfrac{2}{25}\\\\= \dfrac{11}{10}X \dfrac{25}{2}\\\\= \dfrac{11X25}{10X2}\\\\= \dfrac{275}{20}\\\\=13\dfrac{3}{4}$ servings$

Stephen can make $13\dfrac{3}{4}$ servings$ .

You might be interested in

What is repating and terminating

Harlamova29_29 [7]

Repeating decimal is a decimal that repeats once divided...
And, a terminating decimal is a decimal that ends that is it has finite numbers

6 0

3 years ago

Convert y+1=3/4(x-16) to standard form.

Lelu [443]

Standard form: Ax + by =C
so
y+ 1 = 3/4(x - 16)

4(y + 1) = 3(x - 16)
4y + 4 = 3x - 48
3x - 4y = 52

Answer is D. last option
3x - 4y = 52

3 0

3 years ago

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

Simplify completely -4m(9m+5)

harina [27]

Answer:4x

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Jack, Davis, Faiza, and Marry want to take a picture together. In how many ways can they are arranged?

mafiozo [28]

Answer:

16

Step-by-step explanation:

3 0

3 years ago