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rewona [7]
3 years ago
15

A jar contains 8 nickles, 10 dimes, 6 quarters, and 22 pennies. A coin in chosen at random from the jar. What is the probability

that the coin chosen is a dime?
A. 1/23
B.
C. 45
D. 2730
Mathematics
1 answer:
postnew [5]3 years ago
3 0
Total coins=46 There are 10 dimes, so it is 10/46. If you reduce it the answer is 5/23. That is not one of the visible options, but there is no option for B so I could be right!
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17) Convert the number to scientific notation 0.00317
Masteriza [31]

Answer:

3.17x10^-3

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the  

10. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

6 0
3 years ago
Question is on the photo pls help
Bond [772]

Answer:

d)-8

Step-by-step explanation:

  1. -6s-¹t²
  2. -6(3)-¹(-2)²
  3. -6(⅓)(4)
  4. -8
4 0
3 years ago
If \: x + y + z = 0 \: then \: show \: that \: {x}^{3} + {y}^{3 } + {z}^{3} =3xy<br>​
tankabanditka [31]

Answer:

Given

x+y+z=0

⟹x+y=−z

Cubing on both sides

(x+y) 3 =(−z) 3

⟹x 3 +y 3 +3x 2y+3xy 2 =−z 3

⟹x 3 +y 3 +3xy(x+y)=−z 3

⟹x 3+y 3+3xy(−z)=−z 3

⟹x 3 +y 3−3xyz=−z 3

⟹x 3 +y 3 +z 3 =3xyz

Step-by-step explanation:

Hope it is helpful.....

8 0
3 years ago
Please help me solve---- -20-(4x-1)=-15
nikklg [1K]
I hope this helps you -20-(4x-1)=-15 -20-4x+1=-15 -20+1+15=4x -4=4x x=-1
7 0
3 years ago
A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p
Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0
3 years ago
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