How far up a wall will an 11-meter ladder reach, if the foot of the ladder is 4 meters away from the base of the wall?

Which expression represents the perimeter of the rectangle with a width of 4x+3 and legnth of 2x?

slavikrds [6]

P = 2L + 2W
P = 2(2x) + 2(4x + 3)
P = 4x + 8x + 6
P = 12x + 6

The correct response would be D. 12x + 6.

3 0

3 years ago

Find the radius of a circle with an area of 380 square inches. Round your answer to the nearest hundredth

KonstantinChe [14]

Answer:

Step-by-step explanation:

Area of a circle is defined as

A = πr²

Where

A is the area of the circle

r Is the radius of the circle

π is a constant = 3.142

Then,

Given that the area is 380 in²

So,

A = πr²

380 = 3.142 × r²

Divide both side by 3.142

380 / 3.142 = 3.142 × r² / 3.142

111.372 = r²

Take square root of both sides

√111.372 = √r²

10.5533 = r

r = 10.5533 in

Nearest hundredth means to 2d.p

Therefore

r = 10.55 inches

7 0

2 years ago

Read 2 more answers

The length of a rectangle is 4 centimeters less than twice it’s width. The perimeter of the rectangle is 34vm. What are the dime

IrinaVladis [17]

l = 2w - 4

Because we're solving for 2l + 2w, that can be simplified to

2(2w - 4) + 2w = 34

4w - 8 + 2w = 34

6w - 8 = 34

6w = 42

w = 7

Knowing this, we can input w:

2(7) + 2l = 34

14 + 2L = 34

2l = 20

l = 10

L = 10, W = 7, Option C

5 0

3 years ago

Read 2 more answers

Find KL. Round to the nearest hundredth. d Sut of .1.3 48 L

VARVARA [1.3K]

Answer:

1.3

Step-by-step explanation:

give brainliest

4 0

3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago