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3 years ago

Write log7(2 ⋅ 6) + log73 as a single log.

Mathematics

2 answers:

MariettaO [177]3 years ago

6 0

Log7 36 is the correct answer.

Marina86 [1]3 years ago

6 0

Answer:

log₇(2 x 6) + log₇3 = log₇ 36

Step-by-step explanation:

We have the result

log A + log B = log AB

Here the question is log₇(2 x 6) + log₇3.

Using the result we will get,

log₇(2 x 6) + log₇3 = log₇(2 x 6 x 3) =log₇ 36

So, log₇(2 x 6) + log₇3 = log₇ 36

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write five fractions equivalent to 1/2. What is the relationship between the numerator and denominator of fractions equivalent t

Vlada [557]

2/4
6/12
4/8
8/16
9/18
different factions but the same value.

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2 years ago

Quadrilateral ABCD is a rhombus. Find m

bearhunter [10]

Answer:

AC = 18, EB = 5 Step-by-step explanation:In a rhombus, AE = CE, and DE = BE. However, AE or AC ≠ DE or BE.

CE = AE

= 9

AC = CE + AE

= 9 + 9

= 18

DE = BE

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2 years ago

Two friends, Alejandro and Mia, agreed to split the check at a restaurant. Alejandro ordered $12 worth of food and Mia ordered $

kvasek [131]

Cost of Alejandro's food = $12

Tip paid by Alejandro = 20% of his bill = 20% of 12 = 2.4

Cost of Mia's food = $18

Tip paid by Mia = 15% of his bill = 15% of 18 = 2.7

Total Tip Paid = 2.4+2.7 = 5.1

Total Check Amount = 12 + 18 = 30

Total tip paid by Alejandro and Mia is r% of the total check amount.

So, we need to determine what % of 30 is equal to 5.1%

⇒ r% × 30 = 5.1

⇒ r% = 5.1 / 30

⇒ r% = 17%

Hence, r is 17%

8 0

3 years ago

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$