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uranmaximum [27]

3 years ago

13

5.Gabby is fishing from a small boat. Her fishing hook is 36 feet below her, and a fish is swimming at the same depth as the hoo

k, 77 feet away. How far away is Gabby from the fish?

1 answer:

Darina [25.2K]3 years ago

5 0

Answer:

113 feet away

Step-by-step explanation:

This one is just simple maths! 36 + 77.

Hava a great day!

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Which point on the number line indicates a number that is less than 9.1 and greater than 6.5

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Step-by-step explanation:

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3 years ago

Write this number in expanded form in the box below. Use powers of ten in your answer.

yarga [219]

Answer:

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Step-by-step explanation:

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Solve the simultaneous equations by substitution<br> 4x+3y=-2<br> 3y=4-x

stiv31 [10]

Answer: x=-2 y=2

Step-by-step explanation:

I decided that isolating the y value of the second equation is the best way to start solving.

I divide 3 from both sides to get y= (4-x)/3. With this, I can substitute into the first equation. 4x+3(4-x)/3)=-2.

The multiplying the 3 by the denominator three cancels each other out to make:

4x+4-x=-2.

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We end up with 3x=-6. Then, we divide both sides by 3.

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I hope you learned something :)

4 0

3 years ago

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densk [106]

Answer:

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Step-by-step explanation:

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3 years ago

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A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size 'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size n₂ = 24</em>

<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample S₂ = 2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em> </em> $t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

<em> where </em> $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

<em> substitute values and we get S² = 40.988</em>

<em> </em> $t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$ <em></em>

<em> </em> t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em> </em>

<em> </em>

<em> </em>

6 0

3 years ago