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GREYUIT [131]
3 years ago
12

How many strings of 4 decimal digit do not contain the same digit twice?

Mathematics
1 answer:
goblinko [34]3 years ago
7 0
9*9*8*7

You can choose 9 digits for the first digit, 9 Fót the second (because IT can be 0) and 8 Fót the third etc.. hope this helps
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I didn’t study and it’s 2 am pls help
AlexFokin [52]

Answer: 7. g = -50, 8. f= 340/6 9. 3 10 .125

Step-by-step explanation:#7 g(3): 4(3)^2 +6 =12^2 +6 = 1444=6= 150. We divide 150 by 3 giving us -50.  #8We start by putting f(6): -3(6)^2 -4(6) +8. We use PEMDAS. -18^2 -24 +8 = _324-24+8 F= 340/6 #9 g(15) = Square root of 15-6=9 and the square root of 9 is 3, therefore, the answer is 3. #10 h (x) = 2^x we plug -3 to 2^-3. It is a negative number and it gives us .125 or 12.5

3 0
3 years ago
Read 2 more answers
Which statement best describes the polynomial?
amid [387]

Answer:-12x^{2}-9x-4584671418

Step-by-step explanation:

4 0
2 years ago
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You and your friend play a game. You answer 80% of the questions correctly and your friend answers 0.60 of the questions correct
bezimeni [28]

Answer:

5

Step-by-step explanation:

Assuming both players can answer the same question, the minimum number of questions is the smallest number that when multiplied by either 0.60 or 0.80 yields a whole number.

Let x be the number of questions, solving by trial and error:

if\ x=2\\x*0.8=1.6\\x*0.6=1.2\\\\if\ x=3\\x*0.8=2.4\\x*0.6=1.8\\\\if\ x=4\\x*0.8=3.2\\x*0.6=2.4\\\\if\ x=5\\x*0.8=4\\x*0.6=3\\\\

Therefore, the minimum number of questions in the game is 5.

5 0
3 years ago
A random sample of 850 births included 434 boys. Use a 0.10 significance level to test the claim that 51.5​% of newborn babies a
Molodets [167]

Answer:

A

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

z = -0.257

P value = P(Z<-0.257) = 0.797

Decision; we fail to reject the null hypothesis. That is, the results support the belief that 51.5​% of newborn babies are​ boys

Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 90% confidence interval) ---- reject Null hypothesis

Z score < Z(at 90% confidence interval) ------ accept Null hypothesis

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above;

Null hypothesis: H0: p = 0.515

Alternative hypothesis: Ha ≠ 0.515

Given;

n=850 represent the random sample taken

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size = 850

po = Null hypothesized value = 0.515

p^ = Observed proportion = 434/850 = 0.5106

Substituting the values we have

z = (0.5106-0.515)/√(0.515(1-0.515)/850)

z = −0.256677

z = −0.257

To determine the p value (test statistic) at 0.10 significance level, using a two tailed hypothesis.

P value = P(Z<-0.257) = 0.797

Since z at 0.10 significance level is between -1.645 and +1.645 and the z score for the test (z = -0.257) which falls with the region bounded by Z at 0.10 significance level. And also the one-tailed hypothesis P-value is 0.797 which is greater than 0.10. Then we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that at 10% significance level the null hypothesis is valid.

5 0
3 years ago
Equation: (y+12)/(-4)=5<br> What is y?
zlopas [31]
The answer is y= -32
6 0
2 years ago
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