All categories

3 years ago

The length of a rectangle is 4 cm longer than its width. If the perimeter of the rectangle is 44 cm, find its area.

1 answer:

5 0

Length: w+4
Width:w
Perimeter:44
Area: L times w
P=44
44=2(w)+2(w+4)
44=2w+2w+8
44=4w+8
36=4w
w=9
L=w+4
L=9+4
L=13
Area should equal 117 cm squared

You might be interested in

Find the area of this trapezoid. Include the correct unit in your answer.

NemiM [27]

Answer:

$\displaystyle A _{ \text{trapezoid}} = 70 {m}^{2}$

Step-by-step explanation:

we are given a trapezoid

we want to figure out the area

remember that,

$\displaystyle A _{ \text{trapezoid}} = \frac{a + b}{2} h$

where a and b represent the parallel lines and h represents the height

we get from the pic that a and b are 5 and 15 respectively and h is 7

so substitute:

$\displaystyle A _{ \text{trapezoid}} = \frac{5 + 15}{2} \times 7$

simplify addition:

$\displaystyle A _{ \text{trapezoid}} = \frac{20}{2} \times 7$

simplify division:

$\displaystyle A _{ \text{trapezoid}} = 10\times 7$

simplify multiplication:

$\displaystyle A _{ \text{trapezoid}} = 70$

since we multiply two same units we of course have to use square unit

hence,

$\displaystyle A _{ \text{trapezoid}} = 70 {m}^{2}$

4 0

2 years ago

3/5+1/3= ...............................................................................................................

Mrac [35]

Answer:

Step-by-step explanation:

you add two fractions:

3/5+1/3=

find the common denominator = it is 15 (3*5)

9/15+5/15=14/15

7 0

3 years ago

50 POINTS & BRAINLIEST!! PLEASE HELP!!<br><br> (Unrelated answers WILL be reported)

solong [7]

Answer:

a) Trained 730 hours

b) 30.42 days

c) 4.35 weeks

Step-by-step explanation:

Hours the dog trained:

2 * 365 = 730

In days:

730/24 ≈ 30.42

In weeks:

30.42/7 ≈ 4.35

3 0

2 years ago

Thinking about the areas of a Standard Normal Distribution, as z values decrease, do the areas to the left of z increase, or dec

Vera_Pavlovna [14]

The figure below shows the standard normal distribution or "bell-shaped" curve, plotted against the z-score.
The z-score is defined as
z = (x - μ)/σ
where
x = nrandom variable
μ = mean
σ = standard deviation.

As z-values decrease, areas to the left of z decrease as shown by the shaded area.

Answer:
Areas to the left of z decrease.

8 0

3 years ago

Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found

Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let $X_{ij}$ denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

$Xbar_{i}$ = $\frac{1}{J_{i}} summation(X_{ij})$ where J=1 to $J_{i}$ and $J_{i}$ is ith sample size

$s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} - Xbar_{i})^{2}$

The sample sizes are J₁ =12 J₂=10 J₃=18 J₄=9

the total number in all samples combined is 49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

$\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]$

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar= $\frac{1}{49}$ (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ $J_{i} (Xbar_{i} -Xbar)^{2}$ from i=1 to 49

SSTr= 20.9910

$SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}$

$SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}$

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

$F=\frac{MSTr}{MSE}$

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an $F_{3,45}$ distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

$\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]$

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0

3 years ago

The length of a rectangle is 4 cm longer than its width. If the perimeter of the rectangle is 44 cm, find its area.​

The length of a rectangle is 4 cm longer than its width. If the perimeter of the rectangle is 44 cm, find its area.