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3 years ago

8

Kalynn plans to cut an 8 foot sandwich into pieces that are 1/2 foot long. How many pieces of the sandwich will she have after i

t has been cut?

1 answer:

Kazeer [188]3 years ago

3 0

Answer: 16

You divide 8 by 1/2 and you get 16 pieces

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What is the value of x? <br> A) 12 <br> B) 39<br> C) 45<br> D) 33

ICE Princess25 [194]

x^2 = 15^2 + 36^2

X^2 = 225 + 1296

X^2 = 1521

x = sqrt(1521)

x =39

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Please help me find the greatest common factor for these 2 problems..

Kisachek [45]

A greatest common factor is the largest number that goes into two or more numbers (in this case two). To find the GCF of two numbers, we have to find the prime factorization (how to express a number as a product of prime numbers) and then see which numbers are common in both of the prime factorizations.

11. The prime factorization of 12 is 3 * 2 * 2. The prime factorization of 18 is 3 * 3 * 2. Looking at the prime factorizations, we can see that both of them have 3 and 2. That means that the GCF is 3 * 2 which is 6.

12. The prime factorization of 48 is 2 * 2 * 2 * 2 * 3. The prime factorization of 80 is 2 * 2 * 2 * 2 * 5. We see that the numbers shared are 2, 2, 2, and 2. That means that the GCF is 2 * 2 * 2 * 2 or 16.

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What is 3.149 rounded to the hundredths

Dovator [93]

The answer is 3.15 rounded to the hundredths.

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3 years ago

URGENT PLZ HELP FIRST RIGHT ANSWER WILL GET BRAINLYEST. <br> What is the area of this figure

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3 years ago

A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are list

alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Yes the students are reasonably good at estimating one minute

a

Option A is correct

b

The test statistics is $t = 0.354$

Step-by-step explanation:

From the question we are told that

The set of data is

68, 82 , 38 , 62 , 41, 25 , 57 , 64, 67, 47, 61, 71, 91, 87, 64

The population mean is $\mu = 60$

The level of significance is given as $\alpha = 0.01$

The critical value for this level of significance obtained from the normal distribution table is

$Z_{\alpha } = 2.33$

The null hypothesis is

$H_o : \mu = 60 \ seconds$

The alternative hypothesis is

$Ha : \mu \ne 60 \ seconds$

Generally the sample mean is mathematically represented as

$\= x = \frac{\sum x_i }{n }$

where n = 15

So

$\= x = \frac{ 68+ 82 + 38 + 62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}$

$\= x = 61.67$

The standard deviation is mathematically represented as

$\sigma =\sqrt{ \frac{ \sum (x_i - \= x )^2}{n} }$

substituting values

$\sigma =\sqrt{ \frac{ \sum (68-61.67)^2 + ( 82-61.67 )^2 + (38 -61.67)^2 + 62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }$ $\sqrt{ \frac{ ( \cdot \cdot + ( 64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+ (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 + (64-61.67)^2}{15} }$ => $\sigma = 18.23$

The test statistics is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{61.67 -60 }{ \frac{18.23 }{\sqrt{ 15} } }$

$t = 0.354$

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

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3 years ago