All categories

3 years ago

The length of a photograph is 1 less than twice the width. The area is 45 cm squared. Find the dimensions of the photograph.

Mathematics

2 answers:

kkurt [141]3 years ago

7 0

the lenght is 9 and the width is 5 so 9 by 5

Hitman42 [59]3 years ago

4 0

Idk need help as well

You might be interested in

Mr. Coffman asked his students to write an example of a square root with a value greater than 11 but less than 11.5. Select the

Leto [7]

Answer:

Dean and Candace

Step-by-step explanation:

4 0

3 years ago

Two buses leave Houston at the same time and travel in opposite directions. One bus averages 55 mi/h and the other bus averages

Veronika [31]

Well you can just keep adding 55 =1hr until you get to 400 the same with 45

6 0

3 years ago

A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey

Lubov Fominskaja [6]

Answer:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

Step-by-step explanation:

For this case we have the following distribution given:

X 0 1 2 3 4 5 6

P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02

For this case we need to find first the expected value given by:

$E(X) = \sum_{i=1}^n X_i P(X_I)$

And replacing we got:

$E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61$

Now we can find the second moment given by:

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97$

And the variance would be given by:

$Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779$

And the deviation would be:

$Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54$

8 0

3 years ago

Eben, an alien from the planet Tellurango, kicks a football from field level. The equation for the football’s height in meters,

Luden [163]

The maxima of a differential equation can be obtained by getting the 1st derivate dx/dy and equating it to 0.

Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative result in:

dh = - 4 t dt + 12 dt

dh / dt = 0 = - 4 t + 12 calculating for t:

t = -12 / - 4

t = 3 s

Therefore the maximum height obtained is calculated by plugging in the value of t in the given equation.

h = -2 (3)^2 + 12 (3)

h = 18 m

This problem can also be solved graphically by plotting t (x-axis) against h (y-axis). Then assigning values to t and calculate for h and plot it in the graph to see the point in which the peak is obtained. Therefore the answer to this is:

The ball reaches a maximum height of 18 meters. The maximum of h(t) can be found both graphically or algebraically, and lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball to reach maximum height, and the y-coordinate, 18, is the max height in meters.

6 0

3 years ago

Ayliah is 7 years more than 1/2 of Deb's age use x for the variable

irina [24]

Answer:

The expression $\frac{1}{2}x+7$ can be used to find Ayliah's age.

Step-by-step explanation:

Given statement is:

Ayliah is 7 years more than 1/2 of Deb's age.

Let,

x represent Deb's age.

The expression for Ayliah's age would be;

7 more means addition.

Ayliah's age = $\frac{1}{2}x+7$

Hence,

The expression $\frac{1}{2}x+7$ can be used to find Ayliah's age.

7 0

3 years ago