How many polio and measles vaccination did doctor potter give, respectively?

Set questions need immediate help

kompoz [17]

60-35%=25%
50-35%=15%
25%=Math
35%=(mns)
15%=science

4 0

2 years ago

Please help me these right now.

Vladimir79 [104]

49. From 3 coin tosses, there are 8 possible outcomes:

... TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

All but the first have at least one head, so 7/8 of the possibilities have at least one head. That's 87.5% (not among your choices).

Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.

___

Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.

50. See for yourself: the calculator says 66.82%. Your best choice is selection d.

4 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

Don't to mansion!! (omg they actually came)(Not)

dimulka [17.4K]

Answer:

what

Step-by-step explanation: