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4 years ago

Mary is converting 30 yards per hour into feet per hour. What conversion should she use?

Mathematics

1 answer:

Nina [5.8K]4 years ago

6 0

1 yard per hour= 3 feet per hour is the conversion.
Hope this helps

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The final scores of students in a graduate course are distributed normally with a mean of 72 and a standard deviation of 5. What

Tcecarenko [31]

Answer:

0.8041

Step-by-step explanation:

We know that

μ=72 and σ=5

and P(65<X<78)

We can determine the Z value as (X-μ)/σ

P( 65<X<78 )=P( 65-72< X-μ<78-72)

$P((65-72)/5$

$P(65$

To fine the Z values:

$P (-1.4$

From the standard normal tables:

$P (Z$

to find P ( Z<-1.4)

$P ( Z$

From the standard normal tables:

$P ( Z$

Therefore

$P(-1.4$

$P (65$

8 0

3 years ago

The breadth of a rectangular garden is 2/3 of its legth. If its perimeter is 40cm, find its dimensions.

Arisa [49]

Answer:

12; 8

Step-by-step explanation:

length-x

breadth-2/3x

2(x+2/3 x)=40

2×5/3x=40

10/3x=40

x=40÷10/3

x=40×3/10

x=12 (cm) length

2/3×12=8 (cm) breadth

6 0

3 years ago

An angle measuring 3.6 radians is drawn in standard position?

tester [92]

Answer:

III

Step-by-step explanation: I just took the quiz

6 0

3 years ago

A square pyramid has a base length of 6 inches. The height of each triangular face is 11 inches. Which expressions correctly cal

VMariaS [17]

Answer:

168 in²

Step-by-step explanation:

The square would be 36 in²

Area of a triangle is A= 1/2 bh

A = 1/2 (11 x 6), A= 1/2 (66)

1/2 (66) = 33 in², which is the area of a triangle.

Then you multiply 33 x 4 to get the area of all the triangles, which is 132 in²

You add the area of the square to the area of the triangles, which is 168 in².

7 0

3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$

Comparing equations (A) and (B), we have

$5x-1=7x^2-1\\\\\Rightarrow 5x=7x^2\\\\\Rightarrow x=\dfrac{5}{7},~\textup{since }x\neq 0.$

Thus,

$AM=5\times\dfrac{5}{7}=\dfrac{25}{7}=3\dfrac{4}{7}.$

8 0

4 years ago