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Mazyrski [523]
2 years ago
8

If the ratio of raisins to bran flakes in a box of raisin bran flakes cereal is 3:27, how many raisins are there in a box that c

ontains 3,000 raisins and brand flakes?
Mathematics
1 answer:
Jlenok [28]2 years ago
5 0

Answer:

300 raisins

Step-by-step explanation:

sum the parts of the ratio, 3 + 27 = 30 parts

Divide the total by 30 for the value of one part of the ratio.

3000 ÷ 30 = 100 ← value of 1 part of the ratio , then

3 parts = 3 × 100 = 300 ← number of raisins in a box

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A study of one thousand teens found that the number of hours they spend on social networking sites each week is normally distrib
Svetradugi [14.3K]

Given:

Sample Mean <span>= 30<span>
Sample size </span><span><span><span>= 1000</span></span><span>
</span></span></span>Population Standard deviation or <span><span><span>σ<span>=2</span></span><span>
</span></span>Confidence interval </span><span>= 95%</span>

to compute for the confidence interval

Population Mean or <span>μ<span><span>= sample mean ± (</span>z×<span>SE</span>)</span></span>

<span><span>where:</span></span>

<span><span>SE</span>→</span> Standard Error

<span><span>SE</span>=<span>σ<span>√n</span>= 30</span></span>√1000=0.9486

Critical Value of z for 95% confidence interval <span>=1.96</span>

<span>μ<span>=30±<span>(1.96×0.9486)</span></span><span>
</span></span><span>μ<span>=30±1.8594</span></span>

Upper Limit

<span>μ <span>= 30 + 1.8594 = 31.8594</span></span>

Lower Limit

<span>μ <span>= 30 − 1.8594 = <span>28.1406</span></span></span>

<span><span><span>
</span></span></span>

<span><span><span>answer: 28.1406<u<31.8594</span></span></span>

3 0
3 years ago
Larry made 85% on his exam. If there were 140 questions, how many did he answer correctly
ella [17]

Answer:

119

Step-by-step explanation:

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3 0
2 years ago
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal
ohaa [14]

Answer: AAA similarity.


Step-by-step explanation:  CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

and

∠DCE ≅ ∠ACB [same angle]

Hence, by AAA (angle-angle-angle) similarity,

△CED ~ △ABC.

Thus, the correct option is AAA similarity.


8 0
3 years ago
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