All categories

4 years ago

Two more than 5 times a number is 77

Mathematics

2 answers:

disa [49]4 years ago

5 0

X=15 i think i hope that's what u wanted

Aleonysh [2.5K]4 years ago

3 0

2 + 5n = 77
-2 - 2
----------------
5n = 75
--- ---
5 5
n = 15

Your equation is 2+5n=77
Your answer is 15

*I hope this helped :) Let me know if you have any questions

You might be interested in

A sheet of printer paper is 8.5 inches by 11 inches. what is the area of the paper?

Natasha2012 [34]

8.5*11 is 93.5
That is the answer

4 0

4 years ago

Read 2 more answers

Given below are seven observations collected in a regression study on two variables, x (independent variable) and y (dependent v

natka813 [3]

Answer:

Step-by-step explanation:

Hello!

Given the independent variable X and the dependent variable Y (see data in attachment)

The regression equation is

^Y= b₀ + bX

Where

b₀= estimation of the y-intercept

b= estimation of the slope

The formulas to manually calculate both estimations are:

$b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

$b_0= \frac{}{y} - b*\frac{}{x}$

n=7

∑X= 42

∑X²= 292

∑Y= 49

∑Y²= 403

∑XY= 249

$\frac{}{y} = \frac{sumY}{n} = \frac{49}{7} = 7$

$\frac{}{x} = \frac{sumX}{n} = \frac{42}{7} = 6$

$b= \frac{249-\frac{42*49}{7} }{292-\frac{42^2}{7} }= -1.13$

$b_0= 7- (-1.13)*6= 13.75$

^Y= 13.75 - 1.13X

Using the raw data you can calculate the coefficient of determination as:

$R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]}$

$R^2= \frac{(-1.13)^2[292-\frac{(42)^2}{7} ]}{[403-\frac{(49)^2}{7} ]}= 0.84$

This means that 84% of the variability of the dependent variable Y is explained by the response variable X under the model ^Y= 13.75 - 1.13X

I hope this helps!

6 0

4 years ago

Consider circle H below.

hram777 [196]

We have been given a circle H, in which length of arc XY is 78 miles and measure of arc XY is 70 degrees. We are asked to find the radius of circle.

We will use arc length formula to solve our given problem.

$\text{Arc length}=2\pi r\cdot \frac{\theta}{360^{\circ}}$ , where,

r = Radius of circle,

$\theta$ = Central angle corresponding to arc.

We know that the measure of central angle that corresponds to arc XY will be equal to measure of arc XY.

$78=2\cdot 3.14\cdot r\cdot \frac{70^{\circ}}{360^{\circ}}$

$78=2\cdot 3.14\cdot r\cdot \frac{7}{36}$

$78= 3.14\cdot r\cdot \frac{7}{18}$

$78= r\cdot \frac{21.98}{18}$

$r\cdot \frac{21.98}{18}=78$

$r\cdot \frac{21.98}{18}\cdot \frac{18}{21.98}=78\cdot \frac{18}{21.98}$

$r=63.8762511373$

Upon rounding to nearest hundredth, we will get:

$r\approx 63.88$

Therefore, the radius of circle H is approximately 63.88 miles.

7 0

4 years ago

Help i will be giving brainliest!!

Elena L [17]

See the photo attached below, answer is D

4 0

3 years ago

wo balls are chosen randomly from an um containing 8 white, 4 black,and 2 orange balls. Suppose that we win $2 for each black ba

umka21 [38]

Answer:

The probability distribution is shown below.

Step-by-step explanation:

The urn consists of 8 white (W), 4 black (B) and 2 orange (O) balls.

The winning and losing criteria are:

Win $2 for each black ball selected.
Lose $1 for each white ball selected.

There are 8 + 4 + 2 = 14 balls in the urn.

The number of ways to select two balls is, ${14\choose 2}=91$ ways.

The distribution of amount won or lost is as follows:

Outcomes: WW WO WB BB BO OO

X: -2 -1 1 4 2 0

Compute the probability of selecting 2 white balls as follows:

The number of ways to select 2 white balls is, ${8\choose 2}=28$ ways.

The probability of WW is,

$P(WW)=\frac{n(WW)}{N}=\frac{28}{91}=0.3077$

Compute the probability of selecting 1 white ball and 1 orange ball as follows:

The number of ways to select 1 white ball and 1 orange ball is, ${8\choose 1}\times {2\choose 1}=16$ ways.

The probability of WO is,

$P(WO)=\frac{n(WO)}{N}=\frac{16}{91}=0.1758$

Compute the probability of selecting 1 white ball and 1 black ball as follows:

The number of ways to select 1 white ball and 1 black ball is, ${8\choose 1}\times {4\choose 1}=32$ ways.

The probability of WB is,

$P(WB)=\frac{n(WB)}{N}=\frac{32}{91}=0.3516$

Compute the probability of selecting 2 black balls as follows:

The number of ways to select 2 black balls is, ${4\choose 2}=6$ ways.

The probability of BB is,

$P(BB)=\frac{n(BB)}{N}=\frac{6}{91}=0.0659$

Compute the probability of selecting 1 black ball and 1 orange ball as follows:

The number of ways to select 1 black ball and 1 orange ball is, ${4\choose 1}\times {2\choose 1}=8$ ways.

The probability of BO is,

$P(BO)=\frac{n(BO)}{N}=\frac{8}{91}=0.0879$

Compute the probability of selecting 2 orange balls as follows:

The number of ways to select 2 orange balls is, ${2\choose 2}=1$ ways.

The probability of OO is,

$P(OO)=\frac{n(OO)}{N}=\frac{1}{91}=0.0110$

The probability distribution of X is:

Outcomes: WW WO WB BB BO OO

X: -2 -1 1 4 2 0

P (X): 0.3077 0.1758 0.3516 0.0659 0.0879 0.0110

3 0

4 years ago