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Anvisha [2.4K]
3 years ago
13

9 1/3+3(5)-10= Answer quick please

Mathematics
2 answers:
Vikki [24]3 years ago
6 0
The answer is to your question is 5u1/3
chubhunter [2.5K]3 years ago
6 0
The answer to your question is ........

9 1/3+3(5)-10=?
9 times 1/3=3
3 times 5=15
3+15=18
18-10=8
?=8

SO the answer to your question is 8.

I hope this helps you!!
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Side length of cube with volume of 729 ft
navik [9.2K]
The answer is 9, hope this helped <3
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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
What two equivalent fraction for 1/5​
Julli [10]

Answer:

2 over 10 and 4 over 20

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Hey guys I need your help on this!!
kogti [31]

Answer:

Domain: (-∞, ∞) or All Real Numbers

Range: (0, ∞)

Asymptote: y = 0

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

Step-by-step explanation:

The domain is talking about the x values, so where is x defined on this graph? That would be from -∞ to ∞, since the graph goes infinitely in both directions.

The range is from 0 to ∞. This where all values of y are defined.

An asymptote is where the graph cannot cross a certain point/invisible line. A y = 0, this is the case because it is infinitely approaching zero, without actually crossing. At first, I thought that x = 2 would also be an asymptote, but it is not, since it is at more of an angle, and if you graphed it further, you could see that it passes through 2.

The last two questions are somewhat easy. It is basically combining the domain and range. However, I like to label the graph the picture attached to help even more.

As x ⇒ -∞, f(x) ⇒ 0

As x ⇒ ∞, f(x) ⇒ ∞

7 0
3 years ago
Name
miv72 [106K]
Hggyuihdsryui um sorry but I dont no
8 0
3 years ago
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