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gregori [183]
2 years ago
7

Write two numbers that multiply to the value on top and add to the value on bottom -6 x -5

Mathematics
1 answer:
Novosadov [1.4K]2 years ago
6 0

ATypically this is something done for factoring. To do that, I use trail and error. Do you need to do it by a system of equations?

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Who knows how to sketch this kind of graph
Fofino [41]

The graph of f(x) = 0.5sin(4x)  is sketched below

<h3>Graph of trigonometric functions </h3>

From the question, we are to sketch the graph of the given trigonometric function.

The given trigonometric function is

f(x) = 0.5sin(4x)

The graph of f(x) = 0.5sin(4x)  is sketched below

Learn more on Graph of trigonometric functions here: brainly.com/question/4987799

#SPJ1

8 0
2 years ago
The number 16 has two square roots: 4 and
maw [93]

Answer:

4 and -4 are both the square roots of 16

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
If (3a + 3b + 1) (3a + 3b - 1) = 80, then a + b =
seropon [69]

Answer:

a+b = 4\sqrt{3}

Step-by-step explanation:

3(a + b)^2 = 81

(a + b)^2 = 27

a + b = sq rt 27

a + b = \sqrt{9}· \sqrt{3}

a + b = 3\sqrt{3}

8 0
3 years ago
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorab
Katyanochek1 [597]

Answer:

(a) Customer will not purchase the light bulbs at significance level of 0.05

(b) Customer will purchase the light bulbs at significance level of 0.01 .

Step-by-step explanation:

We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected,  and the following information obtained:

Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.

Let Null hypothesis, H_0 : \mu = 750 {means that the true average lifetime is same as what is advertised}

Alternate Hypothesis, H_1 : \mu < 750 {means that the true average lifetime is smaller than what is advertised}

Now, the test statistics is given by;

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

where, X bar = sample mean = 738.44 hours

               s  = sample standard deviation = 38.2 hours

               n = sample size = 50

So, test statistics = \frac{738.44-750}{\frac{38.2}{\sqrt{50} } } ~ t_4_9

                            = -2.14

(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.

(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.

6 0
3 years ago
What are the solutions to the equation x² + 16x + 14 = 0?
poizon [28]

Answer:

-8+5√2 and -8-5√2

Step-by-step explanation:

Given the expression x² + 16x + 14 = 0

USing the general formulas

x = -16±√16²-4(14)/2

x = -16±√256-56/2

x = -16±√200/2

x = -16±10√2/2

x = -8±5√2

Hence the required solutions are -8+5√2 and -8-5√2

3 0
3 years ago
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