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3 years ago

What is the multiplicative rate of change of the function?

Mathematics

1 answer:

BartSMP [9]3 years ago

8 0

Answer:

We know that the multiplicative rate of change of a function is the number by which each next term of an exponential function is increasing or decreasing. We can find multiplicative rate of change by dividing any term of the function by its previous term.

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Solve the system of equations.

dolphi86 [110]

Answer:

the answer is b

b works for each equation

Step-by-step explanation:

4 0

3 years ago

Calculate the side lengths a and b to two decimal places.

Zinaida [17]

Answer:

$a=10.92 units$

$b=14.51 units$

Step-by-step explanation:

We are given that c=6 units

$\angle A=43^{\circ}$

$\angle B=115^{\circ}$

We have to find the side length a and b.

We know that sum of angles of a triangle =180 degrees

$\angle A+\angle B+\angle C=180$

Substitute the values then we get

$43+115+\angle C=180$

$158+\angle C=180$

$\angle C=180-158=22^{\circ}$

sine law: $\frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sin C}$

$\frac{a}{sin 43^{\circ}}=\frac{6}{sin 22^{\circ}}$

$a=\frac{6}{sin 22^{\circ}}\times sin 43^{\circ}$

$a=10.92 units$

$\frac{b}{sin 115^{\circ}}=\frac{6}{sin 22^{\circ}}$

$b=\frac{6}{sin 22^{\circ}}\times sin 115^{\circ}=14.51$

$b=14.51 units$

6 0

3 years ago

A manufacturer of running shoes knows that the average lifetime for a particular model of shoes is 15 months. Someone in the res

AveGali [126]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to show that the designer's claim of a better shoe is supported by the trial results.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 15 \ months$

The sample size is n = 25

The sample mean is $\= x = 17 \ months$

The standard deviation is $s = 5.5 \ months$

Let assume the level of significance of this test is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 15$

The alternative hypothesis is $H_a : \mu > 17$

Generally the degree of freedom is mathematically represented as

$df = n -1$

=> $df = 25 -1$

=> $df = 24$

Generally the test statistics is mathematically represented as

$t = \frac{\= x- \mu }{\frac{s}{\sqrt{n} } }$

=> $t = \frac{ 17 - 15 }{\frac{5.5}{\sqrt{25} } }$

=> $t = 1.8182$

Generally from the student t distribution table the probability of obtaining $t = 1.8182$ to the right of the curve at a degree of freedom of $df = 24$ is

$p-value = P(t > 0.18182 ) = 0.4286$