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3 years ago

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Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. f(x)

= 3(4)(2x-4). h(x) = (x + 2)3 + 1.
A) x=-3
B) x=0
C) x=2
D) x= 4

2 answers:

olasank [31]3 years ago

6 0

<span>Use a table of function values to approximate an x-value in which the exponential function exceeds the polynomial function. f(x) = 3(4)(2x-4). h(x) = (x + 2)3 + 1.
</span><span>
D) x= 4</span>

Kay [80]3 years ago

3 0

Answer:

The correct option is D.

Step-by-step explanation:

The exponential function is

$f(x)=3(4)^{2x-4}$

The polynomial function is

$h(x)=(x+2)^3+1$

Find the values of both functions at x=-3, 0, 2, 4.

x values f(x) h(x)

-3 0.0000029 0

0 0.012 9

2 3 65

4 768 217

From this table we can say that the exponential function f(x) exceeds the polynomial function g(x) at x=4.

Therefore the correct option is D.

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An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58

Olegator [25]

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, $\mu_1$ = true mean voltage for type K batteries.

$\mu_2$ = true mean voltage for type Q batteries.

So, Null Hypothesis, $H_0$ : $\mu_1 = \mu_2$ {mean voltage for these two types of

batteries is same}

Alternate Hypothesis, $H_1$ : $\mu_1 \neq \mu_2$ {mean voltage for these two types of

batteries is different]

<em>The test statistics we use here will be :</em>

$\frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ follows $t_n__1 + n_2 -2$

where, $X_1bar$ = 8.54 and $X_2bar$ = 8.69

$s_1$ = 0.225 and $s_2$ = 0.725

$n_1$ = 37 and $n_2$ = 58

$s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} } = \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2} }{37+58-2} }$ = 0.585 Here, we use t test statistics because we know nothing about population standard deviations.

Test statistics = $\frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58} } }$ follows $t_9_3$

= -1.219

<em>At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.</em>

Therefore, we conclude that mean voltage for these two types of batteries is same.

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3 years ago

Consider a left-tailed test, where the p-value is found to be 0.10. If the sample size n for this test is 49, then the t statist

tigry1 [53]

Answer:

$df=n-1=49-1=48$

We know that the p value is 0.1, and since is a left tailed test would be given by:

$p_v = P(t_{48} < t_{calculated})=0.1$

And we can find the calculated value using excel or the t table.

If we use the following excel code we got: "=T.INV(0.1,48)"

And the calculated value is $t_{calculated}=-1.29944$

Step-by-step explanation:

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the true mean is lower than specified valu $\mu_o$ , the system of hypothesis would be:

Null hypothesis: $\mu \geq \mu_o$

Alternative hypothesis: $\mu < \mu_o$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can find the degrees of freedom for this test given by:

$df=n-1=49-1=48$

We know that the p value is 0.1, and since is a left tailed test would be given by:

$p_v = P(t_{48} < t_{calculated})=0.1$

And we can find the calculated value using excel or the t table.

If we use the following excel code we got: "=T.INV(0.1,48)"

And the calculated value is $t_{calculated}=-1.29944$

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