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3 years ago

Please help!!!

Mathematics

1 answer:

lord [1]3 years ago

7 0

Answer:

1a) Yes

1b) No

1c) Yes

1d) No

1e) Yes

1f) Yes

Step-by-step explanation:

1a) 3+4>5

7>5 Yes

4+5>3

9>3 Yes

5+3>4

8>4 Yes

Yes

1b) 1+4 =5

No

1c) 1+5>5 Yes

5+5>1 Yes

1+5>5 Yes

Yes

1d) 4+3<8 No

No

1e) 8+8>4 Yes

8+4>8 Yes

Yes

1f) 4+4>4 Yes

4+4>4 Yes

Yes

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daniel's basketball team won 23 games and lost 8 games. Write the ratio of wins to losses in simplest form.

Soloha48 [4]

23:8 is the simplest form. However, you can put 5.75:2 or even 2.86/1 if you are looking for the absolute simplest

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3 years ago

A dog owner wants to use 200 ft of fencing to enclose the greatest possible area for his dog. he wants the fenced area to be rec

Fiesta28 [93]

Perimeter (P) = 2 · Length(L) + 2 · Width (W) → P = 2L + 2W
Solve for either L or W (I am solving for L).
200 - 2W = 2L
(200 - 2W)/2 = L
100 - W = L

Area (A) = Length (L) · Width (W)
= (100 - W) · W
      = 100W - W²

Find the derivative, set it equal to 0, and solve:
dA/dW = 100 - 2W
   0 = 100 - 2W
   W = 50

refer to the equation above for L:
100 - W = L
100 - 50 = L
      50 = L

Dimensions for the maximum Area are 50 ft x 50 ft

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3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

Help on this too pls

kenny6666 [7]

Step-by-step explanation:

Half of 40 is 20 and luckily they already gave us how much she would earn if she worked 20 hours.

So all we have to do is multiply the amount of money she earned when she worked 20 hours by 2 to see how much she earned when she worked 40 hours.

20 hours:165 dollars or 20:165

20 x 2= 40 and 165x2=330

So if Monique worked 40 hours she would earn $330.

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2 years ago

Read 2 more answers

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Alekssandra [29.7K]

True......................

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3 years ago