How to solve 36x+3(34x+9)

One number exceeds another by -1. the sum of the numbers is 5. what are the numbers?

maks197457 [2]

X-y=-1
x+y=5
y=x+1
x+x+1=5
2x+1=5
2x=4
x=2
y=x+1
y=2+1
y=3

4 0

3 years ago

Which shows a correct order to solve this story problem? Kent and Curtis went to the state fair. They had to pay a total of $7.1

KiRa [710]

The problem is asking how much each person will need to pay. Simplifying the problem into an equation with variables (an algorithm) will greatly help you solve it:

S = Sales Tax = $ 7.18 per any purchase
A = Admission Ticket = $ 22.50 entry price for one person (no tax applied)
F = Food = $ 35.50 purchases for two people

We know the cost for one person was: (22.50) + [(35.50/2) + 7.18] =
$ 47.43 per person. Now we can check each method and see which one is the correct algorithm:

Method A)
[2A + (F + 2S)] / 2 = [ (2)(22.50) + [35.50 + (2)(7.18)] ]/ 2 = $47.43
Method A is the correct answer

Method B)
[(2A + (1/2)F + 2S) /2 = [(2)(22.50) + 35.50(1/2) + (2)7.18] / 2 = $38.55
Wrong answer. This method is incorrect because the tax for both tickets bought are not being used in the equation.

Method C)
[(A + F) / 2 ]+ S = [(22.50 + 35.50) / 2 ] + 7.18 = $35.93
Wrong answer. Incorrect Method. The food cost is being reduced to the cost of one person but admission price is set for two people.

6 0

3 years ago

V1-v2\t solve formula for v1

vitfil [10]

Answer:

Uwm am I famous but I'm doin dis for free points sowwy

8 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

It is 185 miles to Fort Worth. If Vang drives

astra-53 [7]

2 hour at 65 mile per hour

65 x 2 = 130

130 miles driven within those 2 hours

185 - 130 = 55

Vang is 55 miles away from Fort Worth

hope this helps

4 0

3 years ago