Use the distributive property to find the value of x.

Abhasra spent half of her weekly allowance at the movies. To earn more money her parents let her clean the windows in the house

never [62]

Answer:

$10

Step-by-step explanation:

x is her weekly allowance.

She spent half her allowance and has ½x remaining.

She earned $7, and now has $12.

12 = ½x+7

x = 10

7 0

2 years ago

A random sample of 49 measurements from one population had a sample mean of 15, with sample standard deviation 5. An independent

kotykmax [81]

Answer:

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

Step-by-step explanation:

$\bar X_{1}=15$ represent the mean for sample 1

$\bar X_{2}=18$ represent the mean for sample 2

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=49$ sample size for the group 2

$n_{2}=64$ sample size for the group 2

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=49+64-2=111$

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$t=\frac{(15-18)-0}{\sqrt{\frac{5^2}{49}+\frac{6^2}{64}}}}=-2.897$

P value

Since is a bilateral test the p value would be:

$p_v =2*P(t_{111}$

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the difference between the two groups is significantly different.

7 0

2 years ago

PLEASE HELP THANKS!!

mafiozo [28]

Answer: a) x = 5 b) x = 12

Step-by-step explanation: For each one you just do the inverse to get it on the other side of the equal sign. For example x + 3 = 8 would be like this :

1. x + 3 = 8

2. x + 3 - 3 = 8 -3

( + 3 - 3 cancels out)

3. x = 5

6 0

2 years ago

Read 2 more answers

Only answer if you're very good at math.Please I keep on posting this but nobody is helping me.

hichkok12 [17]

I believe A. but Major guess tho

3 0

2 years ago

How would you solve a problem like this

RUDIKE [14]

Answer:

ΔGJH ≅ ΔEKF

HL: GH and EF

SAS: FK and JH (or GH and EF)

ASA: ∠JGH and ∠FEK (or ∠EFK and ∠JHG)

ΔGFJ ≅ ΔEKH

SSS: KH and FJ

SAS: ∠KEH and ∠FGJ

Step-by-step explanation:

List whatever angles/sides need to be congruent for the two triangles to be congruent.

Prove ΔGJH ≅ ΔEKF using....

- HL (Hypotenuse + Leg)

We already have two legs that are congruent (EK and GJ), so we just need the hypotenuses (GH and EF) to be equal.

- SAS (Side + Angle + Side)

1 pair of sides (EK and JG) are equal, and m∠EKF = m∠GJH. So we need one more side. You can either use FK and JH or GH and EF.

- ASA (Angle + Side + Angle)

1 pair of angles (∠EKF and ∠GJH) are already given as equal, and 1 pair of sides (EK and GJ) are equal. We just need one more pair of angles. So either ∠JGH and ∠FEK or ∠EFK and ∠JHG.

Prove ΔGFJ ≅ ΔEKH using...

- SSS (Side + Side + Side)

Two pairs of sides (EK + GJ and EH + FG) are equal, so KH and FJ need to be equal.

- SAS (Side + Angle + Side)

FG + EH and KE + GJ are equal. We need to use the angle in between them to use SAS, so ∠KEH and ∠FGJ need to be equal.

7 0

2 years ago