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Bad White [126]
3 years ago
10

Help me please Greatest Common Factor

Mathematics
2 answers:
ra1l [238]3 years ago
6 0

Answer:

Step-by-step explanation:

Let's first look at the coefficients: 10, -15, 20

The GCF of these numbers is 5 (all numbers are divisible by 5, nothing larger).

Next, let's look at the x terms: x^{3}, x^{4}, x^{5}

The GCF of these terms is x^{3} (find the lowest exponent)

Finally, let's look at the y terms: y^{5}, y^{8}, y^{6}

The GCF of these terms is y^{5} (find the lowest exponent)

This gives us a GCF of 5x^{3}y^{5}

Anika [276]3 years ago
5 0

Answer:

5x^3y^5

Step-by-step explanation:

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The ratio of peanuts to almonds in a snack mix is 24 to 15. Simplify the ratio.
Marianna [84]

Answer:

The simplified ratio would be 8 to 5.

Step-by-step explanation:

We can get the simplified ratio by finding the largest number that both sides can be divided by. In this context, it is the number 3 so we divide both sides by 3 to simplify the ratio.

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif
algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

e) p_v =P(t_{118}  

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

\bar X_{O}=2.91 represent the mean for the Orange Coast

\bar X_{C}=2.96 represent the mean for the Coastline

s_{O}=0.05 represent the sample standard deviation for Orange Coast

s_{C}=0.03 represent the sample standard deviation for Coastline

n_{O}=60 sample size for Orange Coast

n_{C}=60 sample size for Coastline

\alpha=0.005 Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis:\mu_{O} \geq \mu_{C}  

Alternative hypothesis:\mu_{O} < \mu_{C}  

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

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

And the degrees of freedom are given by df=n_1 +n_2 -2=60+60-2=118  

Replacing the info we got:

t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64  

Part e

We can calculate the p value with this probability:

p_v =P(t_{118}  

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

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A truck driver drives at a constant rate of 60 miles per hour while on the highway. Which equation represents the distance, d, i
lora16 [44]

Answer:

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Step-by-step explanation:

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A train leaves San Diego at 1:00 pm. A second train leaves the same city in the same direction at 3:00 pm. The second train trav
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Answer:

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Step-by-step explanation:

Let x represent the speed of the first train in mph. Since the second train, is 30 mph faster then the first, therefore the speed of the second train is (x + 30).

The first train leaves at 1:00 pm, therefore at 6:00 pm, the time taken is 5 hours. Therefore the distance covered by the first train at 6:00 pm = x mph * 5 hours = 5x miles

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3x + 90 = 5x

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x = 45

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