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Nataly [62]
3 years ago
15

Complete the square to make the binomial x^2-8x a perfect trinomial

Mathematics
1 answer:
borishaifa [10]3 years ago
3 0
The general Formula looks like this:
x^2 +bx + ( \frac{b}{2})^2
This really only involves 2 steps:
1) Take half of "x" coefficient. ----> -8/2 = -4
2) Square that number.   ---------->  (-4)^2 = 16

Perfect trinomial for this example is then:
x^2 -8x+16
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Are 0.34 and 3/4 equivalent? Please explain your answer In complete sentences.
insens350 [35]
No

when 3 divided by 4 is converted into a decimal, it is 0.75 which does not equal 0.34

0.34 can also be written as 34/100
and 3/4 can be written as 75/100

therefore 0.34 and 3/4 are not equivalent
7 0
3 years ago
Which of the following represents the factorization of the binomial below?
Serggg [28]

Answer:

D

Step-by-step explanation:

x² - 81 ← is a difference of squares and factors in general as

a² - b² = (a + b)(a - b)

Here a = x and b = 9, thus

x² - 81 = (x + 9)(x - 9) → D

4 0
4 years ago
Is 5 a integer or a whole number or a rational number? or all?
madreJ [45]
Integer??? Idk I'm sorry....
5 0
3 years ago
Read 2 more answers
Mitchell is one of the goalies for an ice hockey team. He wants to be goalie in the game today, but that depends on who coaches
seropon [69]

Answer:

49% probability that Mitchell will be a goalie today

Step-by-step explanation:

Coach White coaches with 70% probability. So 70% of the time, there is a 55% probability that Mitchell is the goalie.

The other 100-70 = 30% of the time, 35% probability that Mitchell will be the goalie.

What is the probability that Mitchell will be a goalie today?

P = 0.7*0.55 + 0.3*0.35 = 0.49

49% probability that Mitchell will be a goalie today

8 0
3 years ago
Suppose that 73.2% of all adults with type 2 diabetes also suffer from hypertension. After developing a new drug to treat type 2
denpristay [2]

Answer:

a) Option C is correct.

The requirements have not been met because the population standard deviation is unknown.

The null hypothesis is

H₀: p = 0.732

The alternative hypothesis is

Hₐ: p₀ ≠ 0.732

z-test statistic = -0.98

p-value = 0.327086

The obtained p-value is greater than the significance level at which the test was performed at, hence, we fail to reject the null hypothesis & conclude that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

No significant difference between the population proportion of type 2 diabetics with hypertension while using the new drug and the population proportion of all type 2 diabetics with hypertension.

Step-by-step explanation:

The full complete question is attached to this solution

The only major requirements for using the one sample z-test is that the population is approximately normal at least. And the population standard deviation is known. For this question, the conditions of approximate normality for binomial distribution is satisfied;

np = 718 ≥ 10

And np(1-p) = 1000×0.718×0.282 = 201 ≥ 10

But, no information on the population standard deviation is known. But we can carry on with the test because the sample size is large enough for the p-value obtained from t-test statistic will be approximately equal to the p-value obtained from the z-test statistic.

b) For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

And the alternative hypothesis is that there is significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

Mathematically, the null hypothesis is

H₀: p = 0.732

The alternative hypothesis is

Hₐ: p₀ ≠ 0.732

To do this test, we will use the z-distribution because, the degree of freedom is so large, it is large enough for the p-value obtained from t-test statistic will be approximately equal to the p-value obtained from the z-test statistic.

So, we compute the z-test statistic

z = (x - μ)/σₓ

x = sample proportion of type 2 diabetics with hypertension while using the drug = p =(718/1000) = 0.718

μ = p₀ = proportion of all type 2 diabetics with hypertension = 0.732

σₓ = standard error of the sample proportion = √[p(1-p)/n]

where n = Sample size = 1000

p = 0.718

σₓ = √[0.718×0.282/1000] = 0.0142294062 = 0.01423

z = (0.718 - 0.732) ÷ 0.01423

z = -0.984 = -0.98

checking the tables for the p-value of this z-statistic

Note that this test is a two-tailed test because we're checking in both directions, hence the not equal to sign, (≠) in the alternative hypothesis.

p-value (for z = -0.98, at 0.01 significance level, with a two tailed condition) = 0.327086

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 1% = 0.01

p-value = 0.327086

0.327086 > 0.01

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & conclude that there is no significant evidence that the proportion of type 2 diabetics that have hypertension while taking the new drug is different from the proportion of all type 2 diabetics who have hypertension.

Hope this Helps!!!

3 0
3 years ago
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