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il63 [147K]
3 years ago
10

Can someone please help me answer these questions

Mathematics
1 answer:
Fittoniya [83]3 years ago
7 0

Answer:

sue had the busiest day of all

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Express the angle in radian measure. 60°
Strike441 [17]

Pi radians =180 degrees:

<span>That is, π rad = 180°
</span>
60° in radian measure = 60° × π / 180° 

60° × π / 180° = 1/3 π rad
= 0.33333333333π rad 
= 1.0471975512 rad
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3 years ago
A sample survey was taken of 100 students to determine which of three homecoming themes students prefer
docker41 [41]

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c

Step-by-step explanation:

4 white marbles, 6 yellow marbles, 9 green marbles

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The coordinates of the vertices of △GHI are G(−7,2), H(−2,2), and I(−2,8). Find the side lengths to the nearest hundredth and th
adelina 88 [10]

Answer:

GH = 5, HI = 6, GI ≈ 7.81

m∠H = 90°, m∠G ≈ 50°, m∠I ≈ 40°

Step-by-step explanation:

4 0
3 years ago
Y=<br> 4<br> 5<br> x+1<br> y=<br> –<br> 3<br> 5<br> x–6
aleksandrvk [35]

Answer:

   

Step-by-step explanation:

8 0
2 years ago
The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci
Dmitry [639]

Answer:

The test statistic

Z =  1.149

Since the calculated value of Z =  1.149 is less than 1.96 at 5% (0.05) level of significance.

The null hypothesis is accepted

Hence the proportion is not equal 0.04

<u>Step-by-step explanation:</u>

Given data a random sample of 300 circuits is tested, revealing 16 defectives.

The proportion of success

                                     p = \frac{16}{300} =0.053

  Null hypothesis:- H₀ = P ≠0.04

Alternative hypothesis:- H₁ = P =0.04

Q = 1-P = 1-0.04=0.96

Level of significance ∝ =0.05

The test statistic

         Z= \frac{p-P}{\sqrt{\frac{PQ}{n} } }

now substitute all values, we get

Z= \frac{0.053-0.04}{\sqrt{\frac{0.0384}{300} } }

on calculation, Z =  1.149

Since the calculated value of Z =  1.149 is less than 1.96 at 5% (0.05) level of significance.

The null hypothesis is accepted .

Hence the proportion is not equal 0.04

3 0
3 years ago
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