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Taya2010 [7]
2 years ago
14

After working 80 hours at his job, Tyshawn earned a total of $1080. How much does Tyshawn earn

Mathematics
1 answer:
bogdanovich [222]2 years ago
7 0

The amount  Tyshawn earn per hour is $13.5

<h3>Ratio and proportion</h3>

Fractions are written as a ratio of two integers

If after working 80 hours at his job, Tyshawn earned a total of $1080, then;

80 hrs = $1080

In order to calculate the amount earned per hour

Amount per hour = 1080/80

Amount per hour = $13.5

Hence the amount  Tyshawn earn per hour is $13.5

Learn more on ratio here: brainly.com/question/25927869

#SPJ1

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Answer:

x = 3

Step-by-step explanation:

we know that if we have an original line and we are finding the perpendicular we know that the slope of the second line is gonna be the negative reciprocal of the first line and given that we know it must be  perpendicular to the original line it will be heading downwards on the y axis with no slope so instead of y equals it will be x equals and we know that our x value of the point given is three so the answer is x = 3

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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:

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Thanks for letting me help!!

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