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DedPeter [7]
3 years ago
15

Can someone solve this

Mathematics
1 answer:
monitta3 years ago
6 0
Answers would be x = -6 , x = -2/3
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23,600.

Step-by-step explanation:

23|6

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Twice one number is 15 less than a second number. when 13 is added to the second number,the result is 7 less than 9 times the fi
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2 years ago
What is two hundred sixty-five thousandths in standard form?
Debora [2.8K]

Answer:

10^-1*2.65

Step-by-step explanation:

two hundred sixty-five thousandths= 0.265

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7 0
3 years ago
Use the sample information 11formula13.mml = 37, σ = 5, n = 15 to calculate the following confidence intervals for μ assuming th
bija089 [108]

Answer & Step-by-step explanation:

The confidence interval formula is:

I (1-alpha) (μ)= mean+- [(Z(alpha/2))* σ/sqrt(n)]

alpha= is the proposition of the distribution tails that are outside the confidence interval. In this case, 10% because 100-90%

σ= standard deviation. In this case 5

mean= 37

n= number of observations. In this case, 15

(a)

Z(alpha/2)= is the critical value of the standardized normal distribution. The critical valu for z(5%) is 1.645

Then, the confidence interval (90%):

I 90%(μ)= 37+- [1.645*(5/sqrt(15))]

I 90%(μ)= 37+- [2.1236]

I 90%(μ)= [37-2.1236;37+2.1236]

I 90%(μ)= [34.8764;39.1236]

(b)

Z(alpha/2)= Z(2.5%)= 1.96

Then, the confidence interval (90%):

I 95%(μ)= 37+- [1.96*(5/sqrt(15)) ]

I 95%(μ)= 37+- [2.5303]

I 95%(μ)= [37-2.5303;37+2.5303]

I 95%(μ)= [34.4697;39.5203]

(c)

Z(alpha/2)= Z(0.5%)= 2.5758

Then, the confidence interval (90%):

I 99%(μ)= 37+- [2.5758*(5/sqrt(15))

I 99%(μ)= 37+- [3.3253]

I 99%(μ)= [37-3.3253;37+3.3253]

I 99%(μ)= [33.6747;39.3253]

(d)

C. The interval gets wider as the confidence level increases.

8 0
3 years ago
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