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Sever21 [200]
2 years ago
6

Stephen opens a savings account with an initial deposit of $300.

Mathematics
1 answer:
Ratling [72]2 years ago
6 0
D. Y=250x+300
For example let’s say Stephen has been doing this for 5 months
It would be 250(5)+ 300
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Perimeter of triangle K= 19.0 units (3 s.f.)

Step-by-step explanation:

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3 years ago
2 is 3 more than 5 times a number divided by 4
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8 0
3 years ago
Members of the millennial generation are continuing to be dependent on their parents (either living with or otherwise receiving
Morgarella [4.7K]

Answer:

a)

\bf H_0: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is 0.3

\bf H_a: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is greater than 0.3

b) 34%

c) practically 0

d) Reject the null hypothesis.

Step-by-step explanation:

a)

Since an individual aged 18 to 32 either continues to be dependent on their parents or not, this situation follows a Binomial Distribution and, according to the previous research, the probability p of “success” (depend on their parents) is 0.3 (30%) and the probability of failure q = 0.7

According to the sample, p seems to be 0.34 and q=0.66

To see if we can approximate this distribution with a Normal one, we must check that is not too skewed; this can be done by checking that np ≥ 5 and nq ≥ 5, where n is the sample size (400), which is evident.

<em>We can then, approximate our Binomial with a Normal </em>with mean

\bf np = 400*0.34 = 136

and standard deviation

\bf \sqrt{npq}=\sqrt{400*0.34*0.66}=9.4742

Since in the current research 136 out of 400 individuals (34%) showed to be continuing dependent on their parents:

\bf H_0: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is 0.3

\bf H_a: The mean of adults aged 18 to 32 that continue to be  dependent on their parents is greater than 0.3

So, this is a r<em>ight-tailed hypothesis testing. </em>

b)

According to the sample the proportion of "millennials" that are continuing to be dependent on their parents is 0.34 or 34%

c)

Our level of significance is 0.05, so we are looking for a value \bf Z^* such that the area under the Normal curve to the right of \bf Z^* is ≤ 0.05

This value can be found by using a table or the computer and is \bf Z^*= 1.645

<em>Applying the continuity correction factor (this should be done because we are approximating a discrete distribution (Binomial) with a continuous one (Normal)), we simply add 0.5 to this value and </em>

\bf Z^* corrected is 2.145

Now we compute the z-score corresponding to the sample

\bf z=\frac{\bar x -\mu}{s/\sqrt{n}}

where  

\bf \bar x= mean of the sample

\bf \mu= mean of the null hypothesis

s = standard deviation of the sample

n = size of the sample

The sample z-score is then  

\bf z=\frac{136 - 120}{9.4742/20}=16/0.47341=33.7759

The p-value provided by the sample data would be the area under the Normal curve to the left of 33.7759 which can be considered zero.

d)

Since the z-score provided by the sample falls far to the left of  \bf Z^* we should reject the null hypothesis and propose a new mean of 34%.

7 0
2 years ago
Write the number 16,107,320 in expanded form
yarga [219]
16,107,320=10,000,000+6,000,000+100,000+7,000+300+20


8 0
3 years ago
Read 2 more answers
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