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KengaRu [80]
2 years ago
9

Bruno and Willie need a total of $50,000 to start an accounting business. Bruno has spent a lot of time getting the business set

up, so Willie agrees to contribute $12,000 more than Bruno. How much money will Willie contribute?
Mathematics
1 answer:
Gnom [1K]2 years ago
5 0

The amount contributed by Willie and Bruno is $31000 and $19,000 respectively.

Let amount contributed by Willie be x

Let amount contributed by Bruno be y

Since Willie agreed to contribute $12,000 more than Bruno;

Therefore;

  • x - y = 12000....…....... equation (1)

  • x + y = 50,000...…....... equation (2)

By solving the system of equations simultaneously;

  • 2x = 62000

  • x = 31000

  • and y = 50,000 - 31000

  • y = 19,000

In essence; The amount contributed by Willie and Bruno is $31000 and $19,000 respectively.

Read more:

brainly.com/question/23612222

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The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t 2 + 64t,
coldgirl [10]

Answer:

4 seconds

Step-by-step explanation:

When the ball is on the ground h = 0, hence

- 16t² + 64t = 0 ( solve for t )

factor out - 16t

- 16t(t - 4) = 0

equate each factor to zero and solve for t

- 16t = 0 ⇒ t = 0

t - 4 = 0 ⇒ t = 4

the 0 solution is the height of the ball before being hit and

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6 0
3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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Answer:

Step-by-step explanation:

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Reika [66]

Step-by-step explanation:

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