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iVinArrow [24]
3 years ago
14

Amanda builds a scale model of a bridge that is 460 feet in length. She has to use 6-inch long toothpicks to build this model. T

o build a bridge model that is 512 feet in
length, how long will the toothpicks need to be? Round to the nearest whole unit.
6 inches
7 inches
8 inches
None of these choices are correct.
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Mathematics
1 answer:
Alex_Xolod [135]3 years ago
5 0

Answer:

6 inches

Step-by-step explanation:

460 ÷ 6 = 76.666666.

512 ÷ 76.666666 = 6.672

6.672 rounded to a whole number is 6.

Note:

Pls notify me if my answer is incorrect for the other users that will see this message.

<em>-kiniwih426</em>

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Consider the function f(x) = x² + 10x + 25 for x ≥ -5.<br> What is the value of f-¹(x) when x = 4?
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1 year ago
The midpoint of AB is M(1, 6). If the coordinates of A are (8, 4), what are the coordinates of B?
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Answer:

(-6,8)

Step-by-step explanation:

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4 0
3 years ago
A square window has sides that are 40 inches long. what is the window's perimeter?
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3 0
3 years ago
Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June
Shtirlitz [24]

Answer:

t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569

p_v =P(t_{256}>28.569) \approx 0

So with the p value obtained and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2 +10

Alternative hypothesis: \mu_1 >\mu_2 +10

Or equivalently:

Null hypothesis: \mu_1 - \mu_2 \leq 10

Alternative hypothesis: \mu_1 -\mu_2>10

Our notation on this case :

n_1 =129 represent the sample size for group AISI 1078

n_2 =129 represent the sample size for group AISI 1064

\bar X_1 =123.6 represent the sample mean for the group AISI 1078

\bar X_2 =107.6 represent the sample mean for the group AISI 1064

s_1=2.0 represent the sample standard deviation for group 1 AISI 1078

s_2=1.3 represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569

Now we can calculate the degrees of freedom given by:

df=129+129-2=256

And now we can calculate the p value using the altenative hypothesis:

p_v =P(t_{256}>28.569) \approx 0

So with the p value obtained and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.  

7 0
3 years ago
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