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WINSTONCH [101]
3 years ago
11

Natalie had 64 beads. She used 24 beads to make a bracelet. She used 7/8 of the remaining beads to make a necklace. How many bea

ds does Natalie have left?​
Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0

Answer:

Natalie have left with 5 beads.

Step-by-step explanation:

Given:

Natalie had 64 beads. She used 24 beads to make a bracelet. She used 7/8 of the remaining beads to make a necklace.

Now, to find the beads does Natalie have left.

<em>Total beads Natalie had = 64 beads.</em>

<em>And beads she used to make a bracelet = 24 beads</em>.

So, the remaining beads are = 64-24=40\ beads.

Now,<em> the beads she used to make a necklace</em> = 40 of \frac{7}{8}

                                                                             = 40\times \frac{7}{8}

                                                                             = \frac{280}{8}

                                                                             = 35\ beads  

Hence, the beads Natalie have left = remaining beads - beads used to make necklace.

The\ beads\ Natalie\ have\ left\ =\ 40-35

The\ beads\ Natalie\ have\ left\ =\ 5\ beads.

Therefore, Natalie have left with 5 beads.

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Since x= 12 (0.006461) does not fall in the critical region so we accept our null hypothesis and conclude that the coin is fair.

Step-by-step explanation:

Let p be the probability of heads in a single toss of the coin. Then our null hypothesis that the coin is fair will be formulated as

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The test statistic to be used is number of heads x.

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Heads (x)        Probability (X=x)                        Cumulative     Decumulative

0                        1/16384 (1)             0.000061     0.000061

1                         1/16384  (14)         0.00085             0.000911

2                       1/16384 (91)           0.00555             0.006461

3                       1/16384(364)         0.02222

4                       1/16384(1001)         0.0611

5                       1/16384(2002)       0.122188

6                        1/16384(3003)      0.1833

7                         1/16384(3432)      0.2095

8                        1/16384(3003)       0.1833

9                        1/16384(2002)       0.122188

10                       1/16384(1001)        0.0611

11                       1/16384(364)        0.02222

12                      1/16384(91)            0.00555                             0.006461

13                     1/16384(14)              0.00085                           0.000911

14                       1/16384(1)            0.000061                            0.000061

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∝=  P (X≤0)+P ( X≥14 ) = 0.000061+0.000061= 0.000122

Hence critical region is (X≤0) and ( X≥14)

Computation x= 12

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