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3 years ago

What is the y-intercept of the graph of the line whose equation is y=3/4x+1

Mathematics

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

The correct answer is "1".

Step-by-step explanation:

You need only set x to zero, and solve for y:

$y = \frac{3}{4} x + 1\\y = \frac{3}{4}0 + 1\\y = 1$

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Mkey [24]

9514 1404 393

Answer:

4mn/(3m+12)

Step-by-step explanation:

It is often helpful to factor expressions so that common factors can cancel.

$\dfrac{8n^2}{m^2-16}\times\dfrac{m^2-4m}{6n}=\dfrac{8mn^2(m-4)}{6n(m-4)(m+4)}=\dfrac{4mn}{3(m+4)}\\\\=\boxed{\dfrac{4mn}{3m+12}}$

3 0

2 years ago

Find the mean of the data set. if necessary, round to the nearest tenth. 18 14 8 22 25 8

Elden [556K]

To work out the mean of a data set add all the numbers together and divide them by the number of data items there are.

18 + 14 + 8 + 22 + 25 + 8 = 95

95/6 = 15.8 (that answer is rounded)

6 0

3 years ago

Help me please!!!!!!

Levart [38]

You dived 4,005/16

then you will get your answer

4 0

3 years ago

Read 2 more answers

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$

NOW;

the variance of the refund payment to the couple = 28002.944 - 18539.55

the variance of the refund payment to the couple = 9463.394

7 0

2 years ago

2/5 k - 3/5 + 1/10 k

liraira [26]

Answer:

Step-by-step explanation:

2/ 5 k− 3 /5 k+ 1 /10 k = 2 /5 k+ −3 /5 k+ 1 /10 k

Combine Like Terms:

= 2 /5 k+ −3 /5 k+ 1 /10 k =( 2 /5 k+ −3 /5 k+ 1 /10 k)

= −1 /10 k

Answer:

= −1 /10 k

8 0

3 years ago

Read 2 more answers