1.F(x)=2x+6 find f(20) 2.find x when f(x)=20

What is the value of d in the equation d−1.9=9.1 ?

inn [45]

To get your answer, just add 9.1+1.9. You get 11.

You can double check by replacing 11 with D in the equation.

11-1.9 does equal 9.1 so the answer is correct.

6 0

3 years ago

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Which equations would be related to x - 7 = 38

erik [133]

Answer:

-7+x=38,

Step-by-step explanation:

7 0

2 years ago

Due to a packaging error, 4 cans labeled diet soda were accidentally filled with regular soda and placed in a 12 pack carton of

Sophie [7]

Answer:

Probability that both cans were regular soda = $\frac{1}{11}$

Step-by-step explanation:

Probability = $\frac{Desired outcome}{Total possible outcomes}$

We are given 12 total number of cans; 4 cans have been accidentally filled with diet soda.

Probability that first can is a regular soda:

Outcome that first can is a regular soda will give us the number of regular soda available which are 4

Using formula of probability

Total possible outcomes are, n(total) = 12

Desired outcome: 4 (cans of regular soda)

P(1st can) = $\frac{4}{12}$ = $\frac{1}{3}$

Probability that 2nd can is a regular soda:

<em>As we have already taken a can of regular soda from the pack, the total soda in the pack now 11 and the regular soda left are 3.</em>

Total possible outcomes are, n(total) = 11

Desired outcome: 3 (cans of regular soda as one has already been taken)

P(2nd can) = $\frac{3}{11}$

Probability that both cans are regular soda:

P(both) = P(1st can) × P(2nd can)

= $\frac{1}{3} * \frac{3}{11}$

= $\frac{1}{11}$

3 0

3 years ago

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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 $t_{\alpha/2}= 2.79$

And replacing we got:

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

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

7 0

3 years ago

A tennis shop sold twenty $100 racquets at an 8% profit and ten $80 bags at a 15% loss. The profit on the combined transaction w

Sergeeva-Olga [200]

So first let me assume that the Total profit of the racquets are 8 percentage and not individually.

Now to find the profit we can do

20 x 100= 2000

Now
8=x/2000*100
X= 160

And now for loss
10x80=800

Now

15= x/800*100
X=120

Now profit =

160-120 = 40

The profit was only 40

4 0

2 years ago

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