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Nana76 [90]
3 years ago
12

The Bakhshali Manuscript shows that another method for calculating nonperfect squares was being used in India by 400 CE. Use thi

s method to find the approximate value of . The nearest perfect square that is less than 22 is , whose square root is . Add the square root from step 1 to to get 4.75. Calculate the quantity one-half times the square of divided by the value found in step 2, or 4.75. Subtract the value found in step 3 from the value found in step 2, or 4.75. The approximate value of is .
Mathematics
2 answers:
vovikov84 [41]3 years ago
7 0
<span>The nearest perfect square that is less than 22 is 16, whose square root is 4. 

</span><span>Add the square root from step 1 to 3/4 to get 4.75.

</span>Calculate the quantity one-half times the square of  divided by the value found in step 2, or 4.75. (1/2  * (3/4)^2) <span>÷ 4.75 = 0.06.
</span>
Subtract the value found in step 3 from the value found in step 2, or 4.75.
The approximate value of <span>√22 is 4.69.</span>
sleet_krkn [62]3 years ago
4 0

Answer:

1.  22

2. 4 

3. 0.06.

4.  4.69.

Read more on Brainly.com - brainly.com/question/7481589#readmore

Step-by-step explanation:

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If h(x)= -kx + 14 and h(2)=18, find the value of k
trasher [3.6K]

h(x)= -kx + 14

18 = -k*2 + 14   since x=2

18 = -2k +14

subtract 14 from each side

4 = -2k

divide by -2

-2 =k


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if the weight of a package is multiplied by 5/8 the result is 15 lbs. find the weight of the package.
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Let the weight be d
x*5/8=15
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PLEASE HELP ME!!!!!!!
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Answer: The right answer is 85.0

Step-by-step explanation: I just did it

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3 years ago
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Ro
puteri [66]

Answer:

(a) P(0 ≤ Z ≤ 2.87)=0.498

(b) P(0 ≤ Z ≤ 2)=0.477

(c) P(−2.20 ≤ Z ≤ 0)=0.486

(d) P(−2.20 ≤ Z ≤ 2.20)=0.972

(e) P(Z ≤ 1.01)=0.844

(f) P(−1.95 ≤ Z)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)=0.862

(h) P(1.01 ≤ Z ≤ 2.50)=0.150

(i) P(1.20 ≤ Z)=0.115

(j) P(|Z| ≤ 2.50)=0.988

Step-by-step explanation:

(a) P(0 ≤ Z ≤ 2.87)

In this case, this is equal to the difference between P(z<2.87) and P(z<0). The last term is substracting because is the area under the curve that is included in P(z<2.87) but does not correspond because the other condition is that z>0.

P(0 \leq z \leq 2.87)= P(z

(b) P(0 ≤ Z ≤ 2)

This is the same case as point a.

P(0 \leq z \leq 2)= P(z

(c) P(−2.20 ≤ Z ≤ 0)

This is the same case as point a.

P(-2.2 \leq z \leq 0)= P(z

(d) P(−2.20 ≤ Z ≤ 2.20)

This is the same case as point a.

P(-2.2 \leq z \leq 2.2)= P(z

(e) P(Z ≤ 1.01)

This can be calculated simply as the area under the curve for z from -infinity to z=1.01.

P(z\leq1.01)=0.844

(f) P(−1.95 ≤ Z)

This is best expressed as P(z≥-1.95), and is calculated as the area under the curve that goes from z=-1.95 to infininity.

It also can be calculated, thanks to the symmetry in z=0 of the standard normal distribution, as P(z≥-1.95)=P(z≤1.95).

P(z\geq -1.95)=0.974

(g) P(−1.20 ≤ Z ≤ 2.00)

This is the same case as point a.

P(-1.20 \leq z \leq 2.00)= P(z

(h) P(1.01 ≤ Z ≤ 2.50)

This is the same case as point a.

P(1.01 \leq z \leq 2.50)= P(z

(i) P(1.20 ≤ Z)

This is the same case as point f.

P(z\geq 1.20)=0.115

(j) P(|Z| ≤ 2.50)

In this case, the z is expressed in absolute value. If z is positive, it has to be under 2.5. If z is negative, it means it has to be over -2.5. So this probability is translated to P|Z| < 2.50)=P(-2.5<z<2.5) and then solved from there like in point a.

P(|z|

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(k\circ h)(x)=k(h(x))

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