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Leto [7]
4 years ago
14

A number y increased by 5 is at most 28​

Mathematics
2 answers:
White raven [17]4 years ago
5 0

The answer to your question is this:

Iteru [2.4K]4 years ago
4 0

y + 5\leq 28\\y\leq 23

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9. Three dice are tossed. Find the probability of the specified event.<br> A sum of 5
Marina CMI [18]
3/5 I think sorry if I’m wrong
4 0
3 years ago
Find the slope and the y-intercept of each line from it's equation.<br><br> 3x-y=28
Otrada [13]
<span> 
Slope = 3 
 
x-intercept = 28/3 = 9.33333<span> 

y-intercept = 28/-1 = -28.00000

Hope this helped</span></span>
7 0
3 years ago
If the midpoints of the sides of ∆EFG shown below were connected with line segments, what would be the perimeter of the resultin
masya89 [10]

Answer:

17.5

Step-by-step explanation:

I beleive since we are connecting midpoints you divide each side by 2 and add them all up to get the perimeter of the smaller triangle made with the ratio of 1:2

6 0
3 years ago
Could someone please help? specifically with 1a?
Zarrin [17]
Pair up the terms into separate groups. Then factor each group individually (pull out the GCF). Once that is finished, you factor out the overall GCF to complete the full factorization.

8r^3 - 64r^2 + r - 8
(8r^3 - 64r^2) + (r - 8)
8r^2(r - 8) + (r - 8)
8r^2(r - 8) + 1(r - 8)
(8r^2 + 1)(r - 8)

So the final answer is (8r^2 + 1)(r - 8)
--------------------------------------------------------------
Edit:

Problem 1b) Follow the same basic steps as in part A

28v^3 + 16v^2 - 21v - 12
(28v^3 + 16v^2) + (-21v - 12)
4v^2(7v + 4) + (-21v - 12)
4v^2(7v + 4) - 3(7v + 4)
(4v^2 - 3)(7v + 4)

The answer to part B is (4v^2 - 3)(7v + 4)

--------------------------------------------------------------
Second Edit:

I apologize for the first edit. I misread what you were asking initially. Here is problem 2A. We follow the same basic steps as in 1a) and 1b). You'll need to rearrange terms first

27mz - 12nc + 9mc - 36nz
27mz + 9mc - 12nc - 36nz
(27mz + 9mc) + (-12nc - 36nz)
9m(3z + c) + (-12nc - 36nz)
9m(3z + c) -12n(c + 3z)
9m(3z + c) -12n(3z + c)
(9m - 12n)(3z + c)
3(3m - 4n)(3z + c)
4 0
3 years ago
Based on an indication that mean daily car rental rates may be higher for Boston than for Dallas, a survey of eight car rental c
Elina [12.6K]

Answer:

t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058

The degrees of freedom are:

df=8+9-2=15

And the p value would be:

p_v =P(t_{15}>2.058) =0.0287

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are significantly higher than those in Dallas

Step-by-step explanation:

Data given

n_1 =8 represent the sample size for group Boston

n_2 =9 represent the sample size for group Dallas

\bar X_1 =47 represent the sample mean for the group Boston

\bar X_2 =44 represent the sample mean for the group Dallas

s_1=3 represent the sample standard deviation for group Boston

s_2=3 represent the sample standard deviation for group Dallas

We can assume that we have independent samples from two normal distributions with equal variances and that is:

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

Let the subindex 1 for Boston and 2 for Dallas we want to check the following hypothesis:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

The statistic is given by this formula:

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

Where t follows a t student distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

Replacing we got:

\S^2_p =\frac{(8-1)(3)^2 +(9 -1)(3)^2}{8 +9 -2}=9

And the deviation would be just the square root of the variance:

S_p=3

The statitsic would be:

t=\frac{(47 -44)-(0)}{3\sqrt{\frac{1}{8}+\frac{1}{9}}}=2.058

The degrees of freedom are:

df=8+9-2=15

And the p value would be:

p_v =P(t_{15}>2.058) =0.0287

Since we have a p value lower than the significance level given of 0.05 we can reject the null hypothesis and we can conclude that the true mean for car rental rates in Boston are significantly higher than those in Dallas

5 0
4 years ago
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