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3 years ago

Convert 51 in base 10 into base 5

Mathematics

1 answer:

weeeeeb [17]3 years ago

6 0

i cant put a link but i looked up "51 in base 5"

and got 201

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How many minutes did he/she play in all

Lerok [7]

Add the 3 fractions together.
Mixed fractions: 5 3/4 + 2 2/3 + 10 1/6
Convert to improper fractions and then add. Or simply use a calculator.

Total = 223/12 = 18 7/12

5 0

3 years ago

Read 2 more answers

Emily spent $15 on a magazine and four

LiRa [457]

Answer:

Step-by-step explanation:

$15 a magazine.

To calculate how much each candy bar was, you must do $15-$3 = $12

In total, she spent $12 on candy bars. Therefore you must do $12/4 candy bars= $3 a candy bar.

Hope this helps

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3 years ago

Find the first four terms of this geometric sequence for a1=3 and r=-3

Natali5045456 [20]

$a_1=3;\ r=-3\\\\a_2=a_1r\to a_2=3\cdot(-3)=-9\\\\a_3=a_2r\to a_3=-9\cdot(-3)=27\\\\a_4=a_3r\to a_4=27\cdot(-3)=-81\\\\Answer:\{3;-9;\ 27;-81\}$

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3 years ago

Need the answer for number 7 ASAP!!!!

cupoosta [38]

B is the answer because 60/ 12.50 equals 12/2.5

3 0

3 years ago

Listed below are the lead concentrations in mu ??g/g measured in different traditional medicines. Use a 0.10 0.10 significance l

laiz [17]

Answer:

We conclude that the mean lead concentration for all such medicines is less than 17 mu.

Step-by-step explanation:

We are given below are the lead concentrations in mu;

6.5 18.5 21.5 5.5 8.5 4.5 4.5 17.5 15.5 20

We have to test the claim that the mean lead concentration for all such medicines is less than 17 mu.

Let $\mu$ = mean lead concentration for all such medicines.

So, Null Hypothesis, $H_0$ : $\mu$ = 17 mu {means that mean lead concentration for all such medicines is equal to 17 mu}

Alternate Hypothesis, $H_A$ : $\mu$ < 17 mu {means that the mean lead concentration for all such medicines is less than 17 mu}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample mean lead concentration = $\frac{\sum X}{n}$ = 12.25 mu

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 6.96 mu

n = sample size = 10

So, test statistics = $\frac{12.25 -17}{\frac{6.96}{\sqrt{10}}}$ ~ $t_9$

= -2.16

The value of t test statistics is -2.16.

Now, the P-value of the test statistics is given by the following formula;

P-value = P( $t_9$ < -2.16) = 0.031

Now, at 0.10 significance level the t table gives critical value of -1.383 for left-tailed test. Since our test statistics is less than the critical value of t as -2.16 < -1.383, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean lead concentration for all such medicines is less than 17 mu.

7 0

3 years ago