How do I solve this ?

Given f(x)=10-2x, find f(7)

mylen [45]

Answer:

f(7)=10-14=-4

Answer=-4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Perform the indicated operation -3+14= ?

Zigmanuir [339]

The answer is -3+14=11

5 0

3 years ago

Factor 3y3 – 9y2<br> The factored polynomial is

allochka39001 [22]

Answer:

Factor 3y23y2 out of 3y33y3.

3y2(y)−9y23y2(y)-9y2

Factor 3y23y2 out of −9y2-9y2.

3y2(y)+3y2(−3)3y2(y)+3y2(-3)

Factor 3y23y2 out of 3y2(y)+3y2(−3)3y2(y)+3y2(-3).

3y2(y−3)

Step-by-step explanation:

mark me the brainliest please

4 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod. Grade 1 has 10 observations yielding a sample mean

LuckyWell [14K]

Answer:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

$df = n_1 +n_2 -2 = 10+15-2= 23$

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

Step-by-step explanation:

Data given

$\bar X_1 = 1085$ sample mean for group 1

$\bar X_2 = 1034$ sample mean for group 2

$n_1 = 10$ sample size for group 1

$n_2 = 15$ sample size for group 2

$s_1 = 52$ sample deviation for group 1

$s_2 = 61$ sample deviation for group 2

Solution

We want to check if the two means are equal so then the system of hypothesis are:

Null hypothesis: $\mu_1= \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

And the statistic is given by:

$t = \frac{\bar X_1 -\bar X_2}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

And replacing we got:

$t = \frac{1085-1034}{\sqrt{\frac{52^2}{10} +\frac{61^2}{15}}} = 2.240$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 10+15-2= 23$

And the p value would be:

$p_v = 2*P(t_{23} >2.240) = 0.035$

Since the p value is lower than the significance level we have enough evidence to conclude that the true means are different at 5% of significance

8 0

3 years ago

In a rain forest, it typically rains 290 days out of the year. If it rains today, what is the probability that it will rain tomo

Rom4ik [11]

Hey there!

Since we weren't given any probabilities based on the chance of rain after it rained the day before, we can assume the probability will stay the same. We can disregard "If it rains today...".

If it rains 290 days out of the year (365 days), there will be a 290/365 chance of rain of a particular day. You can divide 290 by 365 to get a 79% chance.

Hope this helped you out! :-)

4 0

3 years ago