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vaieri [72.5K]
3 years ago
9

Lauren is making muffins. Her muffin recipe calls for 3 1 4 cups of flour. If Lauren plans to make 5 batches of the muffin recip

e, use what you know about operating with rational numbers to predict the amount of flour she needs. Justify your prediction.
Mathematics
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

15.14 cups is the answer,  just do 5*3 then put the 14 into place it should be the answer

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The equation y + 3 = 5(x-3)
Dimas [21]

Answer:

Your answer is (0,-18)

Step-by-step explanation:

The Y Intercept is (0, -18) and the X Intercept is (18/5, 0)

6 0
2 years ago
Read 2 more answers
Yup help. help no links plssss
seraphim [82]

Answer:

4.103, 4.13, 4.1, 4.1031

Step-by-step explanation:

4 0
3 years ago
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A certain tennis player makes a successful first serve 6969​% of the time. Suppose the tennis player serves 9090 times in a matc
Wewaii [24]

Answer:

a) 4.387

b) Yes, because np & npq are greater than 10.

c) = 0.017          

Step-by-step explanation:

Give data:

p = 0.69

n = 90

a) a

E(X) = np = 62.1

SD(X) = \sqrt{(np(1-p))}

          =\sqrt{90\times 0.69(1- 0.69)}

          = 4.387

b)

np = 62.1  

q = 1 - p  = 1 - 0.69 = 0.31

npq = 19.251

Yes, because np & npq are greater than 10.

c.

P(X \geq 72   ) = P(X > 71.5) [continuity correction]

=    P(Z> \frac{((71.5-62.1)}{ 4.387})

= P(Z> 2.14 )      

= 1 - P(Z<2.14)              

= 1 - 0.983   (using table)          

= 0.017          

8 0
3 years ago
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
2 years ago
Pleaseeee help meee.
DiKsa [7]

Answer:

Step-by-step explanation:

13. Match it with 54

14. Match it with 576

15. Match it with 70

16. Match it with 6

Hope this helps

(Also, do you want how the answers came up? If so please comment below this and tell me that. I  would be more than happy to write the equations if you want.)

3 0
3 years ago
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