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Aleksandr-060686 [28]
3 years ago
15

the worlds landmass are divided into seven continents. The largest continent in terms of landmass is Asia, representing almost 3

0℅ of the earth's land area. In contrast,the smallest continent is Australia, at about 6℅ of the earth's land area.What

Mathematics
1 answer:
dem82 [27]3 years ago
6 0
A none; B africa C Asia and Africa
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Use distributive law to factor the given expression: 3+21a+15b
Reil [10]

Answer:

3(1+7a+5b)

Step-by-step explanation:

The distributive law states that

a\cdot (b+c)=a\cdot b+a\cdot c.

The expression 3+21a+15b consists of three terms:

  1. 3=3·1;
  2. 21a=3·7a;
  3. 15b=3·5b.

You can see that 3 is the common factor of these three terms, thus

3+21a+15b=3(1+7a+5b).

6 0
3 years ago
Mandy has a 10.2-ounce container of mustard. She uses 130 of the mustard to put on her hot dog each time. How many ounces of mus
geniusboy [140]

Answer: it is 10 over 100

Step-by-step explanation:

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3 years ago
The opening of a bird's nest is in the form of a circle the opening has a diameter of 18 centimeters which measurement is closet
Fittoniya [83]

Answer:

The answer is either 18pi or 56.55

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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

5 0
3 years ago
Solve the following equations x1/3=2 <br> Y-2=9<br> (2s)1/2=9<br> 2n-1=16
Nookie1986 [14]

x \times  \frac{1}{3}  = 2 \\  =  > x = 2 \times 3 \\  =  > x = 6

y - 2 = 9 \\  =  > y = 9 + 2 = 11

(2s) \frac{1}{2}  = 9 \\  =  > s = 9

2n - 1 = 16 \\  =  > 2n = 16 + 1 \\  =  > 2n = 17 \\  =  > n =  \frac{17}{2}  = 8.5

<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />

7 0
3 years ago
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