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tester [92]
3 years ago
13

2^2 x 5^2 is the prime factorization of what number

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
8 0

Answer:

100 is the correct answer.

Step-by-step explanation:

=2×2×5×5

=4×25

=100

Hope this helps you. Have a nice day. Please mark as brainliest. It helps a lot. :)

Anarel [89]3 years ago
6 0

Answer:

2^2 × 5^2

2× 2 × 5×5

4 × 25

100

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Which of the following angles has a reference angle of pi/12
Likurg_2 [28]

Given:

\frac{\pi}{12}

Required:

To calculate reference angle

Explanation:

\frac{19\pi}{12}

Required answer:

Option B

5 0
1 year ago
Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern
Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

                            P.Q. =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~ t__n__1-_n__2-2

where, \bar X_1 = sample mean sales receipts for mail-order sales = $81.70

\bar X_2 = sample mean sales receipts for internet sales = $74.60

s_1 = sample standard deviation for mail-order sales = $18.75

s_2 = sample standard deviation for internet sales = $28.25

n_1 = size of sales receipts for mail-order sales = 7

n_2 = size of sales receipts for internet sales = 11

Also, s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} } = 25.11

<em>Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.</em>

<em />

So, 80% confidence interval for the difference between population means, (\mu_1-\mu_2) is ;

P(-1.337 < t_1_6 < 1.337) = 0.80  {As the critical value of t at 16 degree

                                         of freedom are -1.337 & 1.337 with P = 10%}  

P(-1.337 < \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < 1.337) = 0.80

P( -1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < {(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} < 1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

P( (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } < (\mu_1-\mu_2) < (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ) = 0.80

<u>80% confidence interval for</u> (\mu_1-\mu_2) =

[ (\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } , (\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ]

= [ (81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } , (81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } } ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases (\mu_1-\mu_2) is [-9.132 , 23.332].

4 0
3 years ago
On a true-false quiz, the probability of achieving a perfect score by guessing is halved for each additional question on the qui
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For each question, there is a 1/2 chance at getting the question correct by guessing.

Let's take a scenario to better understand.
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The reason why you multiply is because you need each 1/2 simultaneously for a perfect score, which is an important concept when doing binomial probability later on.

Thus, let's use this knowledge to answer the question.
We need the minimum amount of questions such that the probability is less than 1/10.

We can write an inequality for this:

\frac{1}{10} < \frac{1}{2}^{n}
Now, we need to log both sides to find n.
log_(\frac{1}{2}) \frac{1}{10} < n
n > 3.3219...
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Thus, 4 questions is the minimum number of questions needed.
6 0
3 years ago
8. It takes 15 minutes for one person to play one game of bowling. Predict how many games he could
Natalka [10]

Answer:

56 games

Step-by-step explanation:

1 game in 15 minutes, 4x15=60 minutes (an hour)

so, u get 4 games in one hour. now multiply 4 games by 14 hours to get 56

7 0
3 years ago
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Answer:

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