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yan [13]
4 years ago
15

The value of______ divided by 40.01 is close to 1

Mathematics
1 answer:
guapka [62]4 years ago
7 0

The value of 39.6099 divided by 40.01 is close to 1.

Solution:

Given data:

The value of______ divided by 40.01 is close to 1.

close to 1 means not exactly one which is 0.99.

Let the unknown be taken as x.

x ÷ 40.01 = close to 1

x ÷ 40.01 = 0.9999

$\frac{x}{40.01} =0.99

Do cross multiplication, we get

x =0.99\times40.01

x = 39.6099

Hence the value of 39.6099 divided by 40.01 is close to 1.

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To measure the height of a tree, a surveyor walked a short distance from the tree and found the angle of elevation was 43.9°, th
padilas [110]

Answer:

The height of the tree is H = 77.06 m

Step-by-step explanation:

From Δ ABC

AB = height of the tree

\tan 43.9 = \frac{AB}{BC}

\tan 43.9 = \frac{h}{x}

h = 0.9623 x ------- (1)

From Δ ABD

\tan 37.6 = \frac{h}{20+ x}

h = 0.77 (x + 20) ----- (2)

Equating Equation  1 & 2 we get

0.9623 x =  0.77 (x + 20)

0.9623 x = 0.77 x + 15.4

x (0.1923) = 15.4

x = 80.08 m

Thus the height of the tree is given by

H = 0.9623 x

H = 0.9623 × 80.08

H = 77.06 m

Therefore the height of the tree is H = 77.06 m

4 0
4 years ago
Find the missing length
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Answer:

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Step-by-step explanation:

8 0
3 years ago
20-90x-50x^2 <br> find the vertex
lawyer [7]
I think it’s your answer

4 0
3 years ago
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there arre 221 students on a fiel trip five students ride in a van the rest of the students fill the 6 buses enter an equation t
iren2701 [21]

According to the question statement, the total sum of students is 221, which includes the students that ride on bus and the students that ride in a van.

The students that ride in a van are five, the students that ride on bus are 6 times s which is the product of the number of buses and the number of students that are on each bus.

Write all this information into an equation, this way:

6s+5=221

Now, solve the equation for s to find the number of students that are on each bus.

\begin{gathered} 6s+5=221 \\ 6s=216 \\ s=\frac{216}{6} \\ s=36 \end{gathered}

There are 36 students on each bus.

3 0
2 years ago
Problem 1: Given the following probabilities P(Ac) = 0.45, P(B) = 0.34, and P(B|A) = 0.23. Find the following probabilities:
-BARSIC- [3]

Answer:

P(A)=0.55

P(A and B)=P(A∩B)=0.1265

P(A or B)=P(A∪B)=0.7635

P(A|B)=0.3721

Step-by-step explanation:

P(A')=0.45

P(A)=1-0.45=0.55

P(B∩A)=?

P(B|A)=0.23

P(B|A)=(P(A∩B))/P(A)

0.23=(P(A∩B))/0.55

P(A∩B)=0.23×0.55=0.1265

P(A∪B)=P(A)+P(B)-P(A∩B)

=0.55+0.34-0.1265

=0.7635

P(A|B)=[P(A∩B)]/P(B)=0.1265/0.34 ≈0.3721

8 0
3 years ago
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