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Bess [88]
2 years ago
8

A scale drawing of a rectangular park is 5 inches wide and 7inches long. The actual park is 280 yards long. What is the area of

the actual park, in square yards?
Mathematics
2 answers:
choli [55]2 years ago
7 0

280÷7=40. 280-80 = 200. 200×280=5,600 Length divided by scale length to find the amount that each inch scale size equals 40 yards per inch. 5 inches equals 200 yards Then multiply length times width. 200x280=5,600 yds  

Nadusha1986 [10]2 years ago
4 0

Answer: 56,000 square yards.

Step-by-step explanation:

Given : A scale drawing of a rectangular park is 5 inches wide and 7 inches long.

The actual park is 280 yards long.

Let x be the width of the actual park.

We know that the dimensions of scale drawing and the actual figure are proportional.

\Rightarrow\dfrac{\text{Length in drawing}}{\text{Length in actual park}}=\dfrac{\text{width in drawing}}{\text{width in actual park}}\\\\=\dfrac{7}{280}=\dfrac{5}{x}\\\\\Rightarrow\ x=\dfrac{5\times280}{7}=200

thus , width of actual park = 200 yards

Area of actual park =  Length x width

=280 x 200 = 56,000 square yards.

Hence, area of actual park is 56,000 square yards.

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Sladkaya [172]

Answer:

9.2 ounce bottle cost $6.12 is better buy with unit price of $0.22

Step-by-step explanation:

Unit price of any product is given by

unit price = weight of product/ price of product

Given sizes

A

13.5 ounces bottle cost $2.98

cost of 1 ounce of shampoo for this size = cost price/weight of shampoo

 cost of 1 ounce of shampoo for this size =2.98/13.5 = $0.22

Unit price for this size is $.022

B

9.2 ounce bottle cost $6.12

cost of 1 ounce of shampoo for this size = cost price/weight of shampoo

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Unit price for this size is $0.66

As $0.22 is less than $0.66 thus,

9.2 ounce bottle cost $6.12 is better buy.

6 0
2 years ago
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
Charra [1.4K]

1. Consider square ABCD. You know that  

A_{ABCD}=AD^2=200,

then

AB=BC=CD=AD=10\sqrt{2}.

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.

3. Consider trapezoid BFGC. Its area is

A_{BFGC}=\dfrac{FG+BC}{2}\cdot h, where h is the height of trapezoid and is equal to half of AB. Thus,

A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.

5. Note that angles EGC and CGD are supplementary and

\sin \angle CGD=\sin \angle EGC.

Then

A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.

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8 0
2 years ago
A statistics practitioner in a large university is investigating the factors that affect salary of professors. He wondered if ev
Ulleksa [173]

Answer:

Step-by-step explanation:

Hello!

Given the linear regression of Y: "Annual salary" as a function of X: "Mean score on teaching evaluation" of a population of university professors. It is desired to study whether student evaluations are related to salaries.

The population equation line is

E(Y)= β₀ + β₁X

Using the information of a n= 100 sample, the following data was calculated:

R²= 0.23

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Intercept    25675.5           11393

x                  5321                  2119

The estimated equation is

^Y= 25675.5 + 5321X

Now if the interest is to test if the teaching evaluation affects the proffesor's annual salary, the hypotheses are:

H₀: β = 0

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There are two statistic you can use to make this test, a Student's t or an ANOVA F.

Since you have information about the estimation of  β you can calculate the two tailed t test using the formula:

t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } } ~t_{n-2}

t= \frac{ 5321 - 0 }{\frac{2119}{\sqrt{100} } } = 25.1109

The p-value is two-tailed, and is the probability of getting a value as extreme as the calculated t_{H_0} under the distribution t_{98}

p-value < 0.00001

I hope it helps!

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Quadrilateral ABCD is inscribed in circle P as shown. Which statement is necessarily true? A. `"m"/_A +"m"/_B = "m"/_C + "m"/_D
pogonyaev

Opposite angles of an inscribed quadrilateral are supplementary, so it is true that ...

... B. m∠A + m∠C = m∠B + m∠D . . . . . = 180°

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A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. The radius of the ride is 9.0 meters. Which of th
lana [24]

Answer with explanation:

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a=\frac{dv}{dt}=6 m/s

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Also, in terms of mathematics, line from center that is radius , to point of contact that is tangent line ,which is on the circle are perpendicular to each another.

→r(radius vector) ⊥  a(Acceleration vector).

So, In option A, Radius vector is Perpendicular to acceleration of riders.

4 0
2 years ago
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