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Usimov [2.4K]
3 years ago
7

Can someone plz help me with this one?

Mathematics
1 answer:
DedPeter [7]3 years ago
5 0
B is the answer to the question
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Arrange the summation expressions in increasing order of their values.
dolphi86 [110]
∑ 4 * 5^(i-1) = 4 + 20 + 100 + 500 = 624
∑ 3 * 4^(i-1) = 3 + 12 + 48 + 192 + 768 = 1,023
∑ 5*  6^(i-1) = 5 + 30 = 35
∑ 5^(i-1) = 1 + 5 + 25 + 125 = 156
Answer:
∑ (i=1, 2) 5 * 6^(i-1) < ∑ (i=1, 4) 5^(i-1) < ∑ (i=1, 4) 4 * 5^(i-1) <
< ∑ (i=1, 5) 3 * 4^(i-1)

5 0
3 years ago
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At a school carnival, 33 out of 55 tickets sold were early-admission tickets. What percentage of the tickets were early-admissio
7nadin3 [17]

Answer:

60%

Step-by-step explanation:

33/55 = .6

multiple by 100 to convert the decimal into a percentage

.6 × 100 = 60%

8 0
3 years ago
∆ABC is mapped onto ∆A'B'C' which is then mapped onto ∆A"B"C". ∆ABC and ∆A"B"C" are shown in the diagram.
hram777 [196]

Answer:

1rotation translation dilation

2dilation

3the relationship is the shapes are the same just dilated and in different spots. The shapes are in different spots and are different sizes ut they are the same shape making them related.

Step-by-step explanation:

6 0
3 years ago
72÷p=8 world problems
spayn [35]
<span>72÷p=8

</span>p = 72÷ 8
p = 9

hope it helps
7 0
3 years ago
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The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
3 years ago
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