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3 years ago

The table below shows all of the possible outcomes for rolling two six-sided number cubes.

Mathematics

2 answers:

Ede4ka [16]3 years ago

6 0

Answer: your answer would be 36

Masteriza [31]3 years ago

6 0

The answer to this question is D

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Graph the line that represents this equation Y=-5x+2

defon

Answer:

i cant answer this in brainly because of lack of display but if you plug in y=5x+2 into mathaway it will show you on a graph

Step-by-step explanation:

5 0

3 years ago

Please please help im confused again

timama [110]

Answer:

$A_s=179 \ yd$

Step-by-step explanation:

Area of Circle can be calculated using following formula

$A=\pi r^{2}$

Where r is radius of circle

Area of outer circle (r=11 yd)

$A_1=\pi (11)^{2}\\A_1=379.94 \ yd$

Area of inner circle (r=8 yd)

$A_2=\pi (8)^{2}\\A_2=200.96 \ yd$

Area of shaded region

$A_s=A_1-A_2$

$A_s=379.94-200.96\\A_s=179 \ yd$

3 0

2 years ago

257 whole number,integer number, or rational number

aliya0001 [1]

257 is a whole number, interger, and a rational number.

8 0

2 years ago

Read 2 more answers

in a trivia game Joshua gets the score of -4 and the first round and it got five in the second round what is his total score aft

tankabanditka [31]

Joshua’s total score is 1 cuz -4 + 5 = 1

4 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