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amid [387]
3 years ago
12

What is the GCF of 1683t, 4085, and 68t?? O 4 O 483t O 8 O 8837

Mathematics
1 answer:
dedylja [7]3 years ago
8 0

Answer:I’m pretty sure ( not 100% thou ) the awnser would be A) 4

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miv72 [106K]

xis less than or equal to -3

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3 years ago
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Find the value of x.<br> helpppp
Bezzdna [24]

Answer: 55^{\circ}

Step-by-step explanation:

The measure of the third unknown arc is

360^{\circ}-70^{\circ}-110^{\circ}=180^{\circ}.

So, x=\frac{180^{\circ}-70^{\circ}}{2}=\boxed{55^{\circ}}

8 0
2 years ago
If f(1) = 2 and f(n) = –2f(n − 1) then find the value of f(5).
dybincka [34]

Answer: 2f

Explanation:

f(1)=8

f(2)= 2(8) = 16

f(3) = 2(16) = 32

f(4) = 2(32) = 64

f(5) = 2(64) =128

6 0
2 years ago
A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi
Luden [163]

Answer:

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

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

p_v =P(z>1.768)=0.039  

Step-by-step explanation:

Data given and notation

\bar X=3.4 represent the sample mean  

\sigma=0.4 represent the population deviation for the sample

n=50 sample size  

\mu_o =3.3 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

Compute the test statistic  

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

P value

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

p_v =P(z>1.768)=0.039  

6 0
3 years ago
100- 58.90 is equal to what
Radda [10]

Answer:

41.1

Step-by-step explanation:

use calculator

6 0
2 years ago
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