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Setler79 [48]
3 years ago
7

HElp please i need help in question number 5????

Mathematics
1 answer:
likoan [24]3 years ago
7 0

Answer:

g

Step-by-step explanation:

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1. Find the number in each case.
d1i1m1o1n [39]

Answer:

awan ko sayo magsagot ka ng sarili mo

6 0
4 years ago
On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally dist
KiRa [710]
1) The formula used for determining the confidence interval is
Sample mean +/- Critical value (or z-score corresponding to 90% confidence) * Standard
error of mean
8439 +/- 1.645 * 100/sqrt 25
8439 +/- 1.645 * 20
8439 +/- 32.9
2) The only difference in this case is finding the critical value i.e., z-score corresponding to 92% confidence which is 1.75 approximately
Then the confidence interval is
8439 +/- 1.75 * 100/sqrt 25
8439 +/- 1.75 * 20
8439 +/- 35
Lower limit is 8439 - 35 = 8404
The upper limit is 8439 + 35 = 8474
6 0
3 years ago
What is the equivalent fraction
IgorC [24]

Answer:

4/9

Step-by-step explanation:

Notice the repeating symbol on top of the 0.4, so instead of 4/10, it would be 0.44444, which is 4/9 if you turn it into a fraction.

5 0
4 years ago
Read 2 more answers
Iĺl mark brainlist
N76 [4]

Answer:

C. 16√5

Step-by-step explanation:

a squared plus b squared equals c squared!

7 0
3 years ago
Find the zero of each function and state the multiplicity of each zero. Please show all steps.
vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.


Step-by-step explanation:

1. y=(x+3)^3

y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3

Zero: x=-3 multiplicity 3.


2. y=(x-2)^2 (x-1)

y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.

Zeros: x=2 multiplicity 2; x=1 multiplicity 1


3. y=(2x+3)(x-1)^2

y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0
3 years ago
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