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Levart [38]
3 years ago
7

Kim has fraction 1 over 2 cup of almonds. She uses fraction 1 over 8 cup of almonds to make a batch of pancakes. Part A: How man

y batches of pancakes can Kim make with fraction 1 over 2 cup of almonds? (4 points) Part B: On your own paper, draw a fraction model that shows the total number of batches of pancakes that Kim can make with fraction 1 over 2 cup of almonds. Make sure to label the model. Below, explain your model in detail to describe how this model visually shows the solution for Part A. (6 points)
Mathematics
1 answer:
Serga [27]3 years ago
7 0

Answer:

4

Step-by-step explanation:

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A random sample of 300 circuits generated 13 defectives. Use the data to test the hypothesis Upper H Subscript 0 Baseline colon
____ [38]

Complete Question

A random sample of 300 circuits generated 13 defectives. a. Use the data to test

                            H_o : p = 0.05

Versus

 

                          H_1 : p \ne 0.05

Use α = 0.05. Find the P-value for the test

   

Answer:

The  p-value is  p-value =  0.5949        

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 300

    The number of defective circuits is  k = 13

Generally the sample proportion of defective circuits is mathematically represented as

        \^ p = \frac{k}{n}

=>     \^ p = \frac{13}{300}

=>     \^ p = 0.0433

Generally the standard Error is mathematically represented as

       SE  = \sqrt{\frac{p(1- p)}{n} }

=>     SE  = \sqrt{\frac{0.05(1- 0.05)}{300} }

=>     SE  = 0.0126

Generally the test statistics is mathematically represented as

       z =  \frac{\^ p - p }{SE}

=>     z =  \frac{0.0433 - 0.05 }{0.0126}

=>     z =  -0.5317

From the z table  the area under the normal curve to the left  corresponding to  -0.5317  is

         (P < -0.5317 ) = 0.29747

Generally the p-value is mathematically represented as

       p-value =  2 * P(Z <  -0.5317 )

=>     p-value =  2 * 0.29747

=>     p-value =  0.5949        

     

3 0
3 years ago
Investing money in a retirement fund is part of a plan for achieving
Nimfa-mama [501]
Long term goalsif you think about it, you're doing something for your future self, basically. i hope this helps you
7 0
4 years ago
Read 2 more answers
ZA and ZB are supplementary angles. If mZA = (x + 6) and m
Ronch [10]

Answer:

za=24

Step-by-step explanation:

supplementary angles add up to equal 180

180=7x+x+30+6

step 1 combine like terms

180=8x+36

step 2 subtract 36 from each side

144=8x

step 3 divide each side by 8

x=18

now we just plug in 18 to x in x+6

18+6=24

4 0
2 years ago
Read 2 more answers
In the diagram,
forsale [732]

Answer:

Probability that the measure of a segment is greater than 3 = 0.6

Step-by-step explanation:

From the given attachment,

AB ≅ BC, AC ≅ CD and AD = 12

Therefore, AC ≅ CD = \frac{1}{2}(\text{AD})

                                  = 6 units

Since AC ≅ CD

AB + BC ≅ CD

2(AB) = 6

AB = 3 units

Now we have measurements of the segments as,

AB = BC = 3 units

AC = CD = 6 units

AD = 12 units

Total number of segments = 5

Length of segments more than 3 = 3

Probability to pick a segment measuring greater than 3,

= \frac{\text{Total number of segments measuring greater than 3}}{Total number of segments}

= \frac{3}{5}

= 0.6

4 0
3 years ago
If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar
sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers  ( h, i , j , m ), are there with  n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0
3 years ago
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