Evaluate: ab for a = 2 and b = 5

does anyone know the questions and answers for FLVS MATH 7 grade DBA 3.00 module??? someone please help instead of telling me to

Mariana [72]

Answer: hi

Step-by-step explanation:

hi

6 0

3 years ago

2. Determine if either of the following equations are functions? Draw the graphs and explain how

nekit [7.7K]

Answer:

Both equation represent functions

Step-by-step explanation:

The function is the relation that for each input, there is only one output.

A. Consider the equation

$y=-\dfrac{3}{5}x+2$

This equation represents the function, because for each input value x, there is exactly one output value y.

To check whether the equation represents a function, you can use vertical line test. If all vertical lines intersect the graph of the function in one point, then the equation represents the function.

When you intersect the graph of the function $y=-\dfrac{3}{5}x+2$ with vertical lines, there will be only one point of intersection (see blue graph in attached diagram). So this equation represents the function.

B. Consider the equation

$y=x-x^2$

This equation represents the function, because for each input value x, there is exactly one output value y.

When you intersect the graph of the function $y=x-x^2$ with vertical lines, there will be only one point of intersection (see green graph in attached diagram). So this equation represents the function.

4 0

3 years ago

Not sure if u can do it without a protractor but please help

lana [24]

Answer:

i dont know the other questions but i do know the last one and the answer is rhombus.

8 0

3 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

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

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

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

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

4 years ago

Smokers: Out of 780 smokers, 376 have been divorced.

beks73 [17]

For statictics of Out of 780 smokers, 376 have been divorced, Non-smokers: Out of 2855 non-smokers, 902 have been divorced, the 95% confidence interval for smokers and non-smokers is mathematically given as

95% confidence interval = (0.5352, 0.4462)
95% confidence interval = (0.3424, 0.2894)
53% increased risk of divorce for smokers.

<h3>What is the 95% confidence interval for smokers and non-smokers?</h3>

Generally, the equation for the Confidence interval is mathematically given as

p ± Z/2[p(1-p)]/n

Where

Z1/2=1-(0.05/2)

Z1/2=0.975

Read z table we have

Z score= 1.96

Hence

0.4907 ± 1.96 (0.4907)(1-0.4907)/485

0.4907±0.0445

Therefore

95% confidence interval = (0.5352, 0.4462)

b)

Z1/2 = 1- (0.05/2)

Z1/2 = 0.975

Z score= 1.96

0.3159 ± 1.96 (0.3159)(1-0.3159)/1184

0.3159±0.0265

Thereofore

95% confidence interval = (0.3424, 0.2894)

c)

In conclusion, The 95% confidence interval helps us read that 53% increased risk of divorce for smokers.