A sample is termed ____ if it has similar characteristics to the population being studied.

3x + 6 = y<br><br> solve for x

tekilochka [14]

Answer:

x=1/3y - 2

Step-by-step explanation:

8 0

3 years ago

0.6 divided by 12.9 show all work

slavikrds [6]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Help and I will follow

inn [45]

Answer:

32

Step-by-step explanation:

Our total is x.

Maysa took half of the cherries, so there is .5x left.

Nabilla took half of the remaining, so that is .5*0.5x, which is .25x

Abdulla took 3.

Your equation is:

.25x-3=5

.25x=8

x=32

So there are 32 cherries

7 0

3 years ago

mariarad [96]

Answer:

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.

Step-by-step explanation:

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

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

$\sigma=2.5$

And the statistic is given by this formula:

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

Where z follows a normal standard distribution

The system of hypothesis on this case are:

Null hypothesis: $\mu_c \leq \mu_t+1$

Alternative hypothesis: $\mu_c >\mu_t+1$

Or equivalently:

Null hypothesis: $\mu_c - \mu_t \leq 1$

Alternative hypothesis: $\mu_c-\mu_t>1$

Our notation on this case :

$n_c =45$ represent the sample size for group control

$n_t =45$ represent the sample size for group treatment

$\bar X_c =5.2$ represent the sample mean for the group control

$\bar X_t =3.1$ represent the sample mean for the group treatment

And now we can calculate the statistic:

$z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087$

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

$p_v =P(z>2.087) =0.0184$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.

5 0

3 years ago

Find 56(−4⋅27). Write your answer in simplest form.

stellarik [79]

Answer:

$-6048$