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2 years ago

Can someone please help me solve this with showing work

1 answer:

6 0

We substitute the numbers first.

(315)/5 - 10(-2^5)

Divide and do the exponent.

70 - 10(-32)

Now multiply 10(-32)

70 - (-320)

Double negative turns to positive.

390

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NEEED HELP FROM BRAINLIEST??????

Leno4ka [110]

Answer:

Step-by-step explanation:

5 0

3 years ago

According to the fundamental theorem of algebra, which polynomial function has exactly 6 roots? f (x) = 5 x superscript 4 baseli

MariettaO [177]

The polynomial with 6 roots is the one of degree 6, so the correct option is:

$f(x) = 7x^6 + 3x^2 + 12$

<h3>Which polynomial function has 6 roots?</h3>

You need to remember that if p(x) is a polynomial of degree N, then it has N roots (such that some of these N roots may be equal or not).

Then we just need to find the polynomial of degree 6. (Remember that the degree represents the maximum exponent of the polynomial).

In this case, the only one of degree 6 is:

$f(x) = 7x^6 + 3x^2 + 12$

Which is the last option.

If you want to learn more about polynomials, you can read:

brainly.com/question/4142886

#SPJ4

4 0

2 years ago

A baseball team is in second place with the wind to loss ratio of .535. what percent of the games played did the team win?

Vsevolod [243]

Answer:

34.85%

Step-by-step explanation:

Let w be the probability of winning, then probability of losing would be (1 - w)

We know the ratio of winning to losing is 0.535

w/(1 - w) = 0.535

w = 0.535 - 0.535w

1.535w = 0.535

w = 0.348534202

% of games won:

0.348534202 × 100

34.8534202%

4 0

3 years ago

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

$s_2=2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

And the deviation would be just the square root of the variance:

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$