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ElenaW [278]
2 years ago
7

20a + 158 +6c = 20 (6) + 15 (4) + 6(8) + ? + ?

Mathematics
1 answer:
Jobisdone [24]2 years ago
7 0

Step-by-step explanation:

20a + 158+6c=you will expand the brackets

= so it is 120+60+48

20a+158+6c=120+60+48

20a+ 6c=120+60+48-158

20a+6c=70

20a divide by 20+6c divide by 6 on both side

a + c = 0.58

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In a bolt factory, machines 1,2, and 3 respectively produce 20%, 30% and 50% of the total output. of their respective outputs 5%
lorasvet [3.4K]

Answer:

a. 1/10 or 10%

b. 1/2 or 50%

Step-by-step explanation:

Since the combination of machines 1, 2 and 3 produce 100% of the total output when added together, then the probability of choosing a bolt at random that is defective is: 5 + 2 + 3 = 10% out of 100% or 10/100, which is 1/10 or 10%.

If the bolt that is choosen at random is defective, than the probability that it came from machine 1 is 5/10 or 1/2 which is also 50%.

3 0
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The two sets of data below show the number of bubbles blown in fifty trials using two different bubble solutions.
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6 0
3 years ago
Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0
Gre4nikov [31]

Answer:

Using trigonometric ratio:

\sec \theta = \frac{1}{\cos \theta}

\tan \theta = \frac{\sin \theta}{\cos \theta}

From the given statement:

\cos \theta = -\frac{4}{5} and sin < 0

⇒\theta lies in the 3rd quadrant.

then;

\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}

Using trigonometry identities:

\sin \theta = \pm \sqrt{1-\cos^2 \theta}

Substitute the given values we have;

\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}

Since, sin < 0

⇒\sin \theta = -\frac{3}{5}

now, find \tan \theta:

\tan \theta = \frac{\sin \theta}{\cos \theta}

Substitute the given values we have;

\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}

Therefore, the exact value of:

(a)

\sec \theta =-\frac{5}{4}

(b)

\tan \theta= \frac{3}{4}

7 0
3 years ago
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