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nexus9112 [7]
3 years ago
13

Together Kate and Matt have 47 cards. If Kate and Zach combine cards they have 126. If Matt and Zach combine they have 93. How m

any cards does Matt have
Mathematics
2 answers:
kenny6666 [7]3 years ago
5 0

Answer: Matt has 7 cards.

Step-by-step explanation:

Let x represent the number of cards that Kate has.

Let y represent the number of cards that Matt has.

Let z represent the number of cards that Zach has.

Together Kate and Matt have 47 cards. This means that

x + y = 47

y = 47 - x

If Kate and Zach combine cards they have 126. This means that

x + z = 126

z = 126 - x

If Matt and Zach combine they have 93. This means that

y + z = 93 - - - - - - - - - 1

Substituting y = 47 - x and z = 126 - x into equation 1, it becomes

47 - x + 126 - x = 93

47 + 126 - x - x = 93

173 - 2x = 93

2x = 173 - 93

2x = 80

x = 80/2 = 40

The number of cards that Matt has is

y = 47 - x = 47 - 40

y = 7

Ede4ka [16]3 years ago
3 0

Answer:

Matt has 7 cards.

Step-by-step explanation:

Let K, M and Z represent number of cards that Kate, Matt and Zach have.

We have been given that together Kate and Matt have 47 cards. We can represent his information in an equation as:

K+M=47...(1)

We are also told that Kate and Zach combine cards they have 126. We can represent his information in an equation as:

K+Z=126...(2)

Further, Matt and Zach combine they have 93. We can represent his information in an equation as:

M+Z=93...(3)

Upon subtracting equation (3) from equation (2), we will get:

K-M=126-93

K-M=33

K=33+M

Upon substituting this value in equation (1), we will get:

33+M+M=47

33-33+2M=47-33

2M=14

M=\frac{14}{2}=7

Therefore, Matt has 7 cards.

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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%.

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