HELP ME WITH THIS PLZ Just write what ones they are plz :)

Can someone please help me I really need help

antoniya [11.8K]

It should be X < 25/52

4 0

3 years ago

Read 2 more answers

2(h-8) -h=h-16 A. 8 B. -8 C. No solution D. Infinitely many solutions

Grace [21]

Answer:

D. Infinitely many solutions

Step-by-step explanation:

The equation simplifies to ...

2h -16 -h = h -16

h -16 = h -16 . . . . . . a "tautology", true for all values of h.

There are infinitely many solutions.

7 0

2 years ago

Are the equations 4-5x=24 and 5x=20 equivalent?

mina [271]

Answer:

not equivalent

Step-by-step explanation:

solve for x

4-5x=24

-5x=20

x = - 4

5x=20

x = 4

no they are not equivalent

8 0

2 years ago

Due Nov 19, 9:10 AM A museum requires that for every 6 students. two teachers are needed. If 42 students are in , ? Participatio

nata0808 [166]

Answer:

too late

Step-by-step explanation:

its dec

3 0

3 years ago

Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

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

The statistic for this case is given by:

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

The pooled variance is:

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

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago