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3 years ago

Please solve and explain steps for this problem.

Mathematics

1 answer:

Helen [10]3 years ago

7 0

This means Square root (^1/2). this is kind of hard to read. anyways. 32^(1/2) separates into 4^(1/2) and 8^(1/2), which further separates into 4^(1/2) and 2^(1/2), root 4 becomes 2, their are two root 4's, so you get 2 x 2, and you are left with 2^(1/2) now why go to all this trouble. because now you can multiply the 4 you created (2x2) times, the 7... giving you 28*2^(1/2) now subtract it from the the other one with root 2. -5*2^(1/2), giving you 23*2^(1/2)-...idk what that last bit is. if its a odd number then this is the end of the problem, if you can get it to root 2. then do that and simplify.

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Answer:

See below ~

Step-by-step explanation:

$\textsf {Each of the steps has been reasoned below :}$

$\implies \textsf {3x - 2 = 4 (Given)}$

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$\textsf {In the third step, the property has been applied by dividing 3 on each side.}$

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Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

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3 years ago

An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing

wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1 $\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

$S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

$S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

$S=0.9905$

Generally the equation for test statistics is mathematically given by

$t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

$t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$t=-4.00091$

Therefore From table

$P value=P(t_{15+15-2}=4.00091)$

$P value=0.00041918$

$P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

3 0

3 years ago