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3 years ago

Select one answer choice For part A and select one answer choice for part B

Mathematics

1 answer:

sweet [91]3 years ago

6 0

$\bf \qquad \qquad \qquad \textit{discriminant of a quadratic} \\\\\\ \stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+6}x\stackrel{\stackrel{c}{\downarrow }}{+9}=0 ~~~~~~~~ \stackrel{discriminant}{b^2-4ac}= \begin{cases} 0&\textit{\underline{one real solution}}~~\textit{\Large \checkmark}\\ positive&\textit{two real solutions}\\ negative&\textit{no solution} \end{cases} \\\\\\ 6^2-4(1)(9)\implies 36-36\implies 0$

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What's 3.318 divided by 0.07? THIS IS HOMEWORK AND IT'S DUE IN 2 DAYS

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The answer is ...............47.4

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3 years ago

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(b) Write the ratio 12:30 in the form 1:n

kari74 [83]

answer:

1 : 2.5

explanation:

given ratio: 12 : 30

divide by 12:

$\frac{12}{12}$ : $\frac{30}{12}$

1 : 2.5

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2 years ago

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2/3 + w = 3/2 what is w

german

Answer:

0.8333

Step-by-step explanation:

2/3 + x

x + 2/3

x + 2/3 = 3/2

x + 2/3 - 3/2 = 0

2/3 - 3/2 = 4/6 - 9/6 = - 5/6

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f(x) - ax + b

f(x) = x - 5/6

a = 1

b = - 5/6

x- 5/6= 0

x = 5/6

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6 0

3 years ago

Which is greater: 25% of 15 or 15% of 25? Explain your reasoning using mathematical evidence or visual models.

d1i1m1o1n [39]

15% of 25 is the greatest of the two.

Answer:

25% of 15 is 3.75

15% of 25 is also 3.75

15 - 3.75 = 11.25

25 - 3.75 =21.25

Compare and then determine.

11.25 < 21.25

8 0

3 years ago

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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