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3 years ago

Describe the journey graphs maths PLZZZZ help

Mathematics

1 answer:

VARVARA [1.3K]3 years ago

8 0

One is going up and then going down and then going up and all the way up. You’re welcome

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Kelly can type 120 words in 3 minutes. at that ate, how many words can she type in 5 minutes?

madam [21]

$\frac{120\ \text{words}}{3\ \text{minutes}} = \frac{40\ \text{words}}{1\ \text{minute}}$

$\frac{40\ \text{words}}{1\ \text{minute}} \times \frac{5}{5} = \frac{200\ \text{words}}{5\ \text{minutes}}$

Kelly can type 200 words in five minutes.

5 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

Jada is solving the equation shown below. Negative one-half (x + 4) = 6 Which is a possible first step to begin to simplify the

elena-14-01-66 [18.8K]

The two options that are the possible first step to begin to simplify the equation are (c) Multiply both sides of the equation by –2. and (d). Distribute Negative one-half over (x + 4).

<h3>How to determine which two options are the possible first step to begin to simplify the equation?</h3>

The equation is given as:

Negative one-half (x + 4) = 6

Rewrite the above equation properly

So, we have

-1/2(x + 4) = 6

A possible first step is to multiply both sides of the equation -1/2(x + 4) = 6 by -2

So, we have

x + 4 = -12

Another possible first step is to distribute the expression -1/2(x + 4) in the equation

So, we have

-1/2x - 1/2 * 4 = 6

Hence, the two options that are the possible first step to begin to simplify the equation are (c) Multiply both sides of the equation by –2. and (d). Distribute Negative one-half over (x + 4).

Read more about equations at:

brainly.com/question/2972832

#SPJ1

3 0

1 year ago

For every 4 miles vanessa jogged cody jogged 3 miles if vanessa jogged 1 miles how far did cody have jogged

Sergio [31]

divide 3 by 4 = 3/4 = 0.75

so Cody jogs 3/4 mile for every 1 mile Vanessa runs

5 0

3 years ago

Alang owns a small business selling ice-cream. He knows that in the last week 62

anzhelika [568]

Step-by-step explanation:

last week there were

62 + 60 + 2 = 124 customers.

that means that the fraction of the customers, who paid by credit card, was

2/124 = 1/62

in situations like that, when we use history data and try to predict future behavior, we use that past experience and extrapolate to a predicted result.

we simply say that we expect the same fraction, 1/62, if customers to use credit cards.

for the absolute number of predicted credit card customers we multiply the predicted number of customers by the expected fraction

1000 × 1/62 = 500/31 = 16.12903226... ≈ 16

so, he expects 16 credit card customers.

5 0

2 years ago