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Oksi-84 [34.3K]
3 years ago
9

Find the area of the triangle. 7 in. 18 in. The area is square inches.​

Mathematics
1 answer:
Klio2033 [76]3 years ago
8 0
63 square inches

(7 x 18) / 2 = 63
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What conclusion can be made based on this equation 8×6=48
Readme [11.4K]

Answer:

That 48 is a multiple of 8 and 6. That 8 and 6 are factors of 48 (which is the same as saying that 8 and 6 are divisors of 48).

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A store has 13 boxes of candles. There are 44
schepotkina [342]

Answer:

145 candles

Step-by-step explanation:

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3 years ago
Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project.
Kobotan [32]

Answer:

t=\frac{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

Step-by-step explanation:

Data given and notation

\bar X_{A}=3.7 represent the mean for athletes  

\bar X_{NA}=4.1 represent the mean for non athletes  

s_{A}=1.1 represent the sample standard deviation for athletes  

s_{NA}=1.3 represent the sample standard deviation for non athletes

n_{A}=32 sample size for the group 2  

n_{NA}=45 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 population mean for athletes is lower than the population mean for non athletes, the system of  hypothesis would be:  

Null hypothesis:\mu_{A}-\mu_{NA}\geq 0  

Alternative hypothesis:\mu_{A} - \mu_{NA}< 0  

We don't have the population standard deviation's, we can apply a t test to compare means, and the statistic is given by:  

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

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{(3.7-4.1)-0}{\sqrt{\frac{1.1^2}{32}+\frac{1.3^2}{45}}}}=-1.457

P value  

We need to find first the degrees of freedom given by:

df=n_A +n_{NA}-2=32+45-2=75

Since is a one left tailed test the p value would be:  

p_v =P(t_{75}  

Comparing the p value with a significance level for example \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't say that the population mean for the athletes is significantly lower than the population mean for non athletes.  

5 0
3 years ago
When and image is dilated, the area of the new image is the [blank] of the [blank] of the original figure and the [blank] factor
finlep [7]

Answer: ummmmmmmmmmm smatrofhdkbdjbd

Step-by-step explanation:

5 0
3 years ago
Please help with answers and showing work
avanturin [10]

Answer:

1:c

let x be the distance

12^2=10^(2) + x^(2)

x=6.6332


2.D

length of 2->3 and 2->1 is 90ft, they are perpendicular

by using Pythagorus theorem

the diagonal of the square = square root of the sum of 90^2 +90^2

=127.279ft


3.B

12^2  -6^2 = 10.39ft


Step-by-step explanation:


4 0
3 years ago
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