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klemol [59]
2 years ago
12

An arch is in the form of a parabola given by the function h = -0.06d^2 + 120, where the origin is at ground level, d meters is

the horizontal distance and h is the height of the arch in meters.
Graph this function on your graphing calculator then complete the following statements.
The height of the arch is: ------- m
The width to the nearest meter, at the base of the arch is ------ m
Mathematics
1 answer:
mylen [45]2 years ago
4 0

Answer:

See attachment for graph

The height of the arch is: 120 m

The width to the nearest meter, at the base of the arch is 22 m

Step-by-step explanation:

Given

h = -0.06d^2 + 120

Solving (a): The graph

See attachment for graph

Variable h is plotted on the vertical axis while variable d is plotted on the horizontal axis.

Solving (b): The height

The curve of h = -0.06d^2 + 120 opens downward. So, the maximum point on the vertical axis represents the height of the arch,

Hence:

height = 120

Solving (c): The width

The curve touches the horizontal axis at two different points.

x_1 = -11

x_2 = 11

The absolute difference of both points represents the width.

So:

Width = |x_2 - x_1|

Width = |11 - -11|

Width = |11 +11|

Width = |22|

Hence:

Width = 22

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Answer:

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2 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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3 years ago
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The probability of picking the third coin as a quarter = 3/8 ( 3 quarters out of 8 coins left).

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3 years ago
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<em>First, find the greatest common factor (GCF) of the numerator and denominator.</em>

<u>Factors of 18</u>: 1, 2, 3, 6, 9, 18
<u>Factors of 24</u>: 1, 2, 3, 4, 6, 8, 12, 24
<u>Common Factors</u>: 1, 2, 3, 6
<u>GCF</u>: 6

<em>Now, divide the numerator by 6 and the denominator by 6.</em>

18 ÷ 6 = 3
24 ÷ 6 = 4

<em>Set these as your new numerator and denominator.</em>

\frac{3}{4}

The answer is (b).

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Answer:

4

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