What is 0.941 repeating as a fraction

A manufacturer of large kitchen appliances keeps track of the costs of warranty claims. Management suspects that 75%

almond37 [142]

Answer:

The correct answer is C.

Step-by-step explanation:

C. ✓ No, the Large Counts condition is not met.

6 0

2 years ago

How do i solve things that dont make sense to me even though i do know but dont know

kirill [66]

Answer:

Just try to make it make sense to you.

Step-by-step explanation:

For instance, a + 1 = 5

a = 4

We know that we should switch the terms as when it comes to normal numbers ,10 + 1 = 11 , 10 = 11-1. Therefore, the only trick is not to memorise, but try to understand the logic behind it. Hope this helps in some way???

7 0

2 years ago

Need help ASAP I dont know what to do please help

shusha [124]

Answer:

the first option

Step-by-step explanation:

equilateral triangles are also equiangular which means each of the three angles is 60°

but you can have an acute triangle with angle measures that are all acute (for example, 59°, 85°, and 36°) that is not equilateral

8 0

3 years ago

Order these numbers from least to greatest. 7.39, 7.3962, 7.309, 7.8°

Hatshy [7]

Answer:

7.8, 7.309, 7.39, 7.3962

Hope this helps and my apologies if this is wrong it should not be though I did it on paper lol.

6 0

3 years ago

Conservationists have despaired over destruction of tropical rain forest by logging, clearing, and burning." These words begin a

IgorLugansk [536]

Answer: -8.16 to 15.84

Step-by-step explanation: Confidence Interval is an interval in which we are a percentage sure the true mean is in the interval.

A confidence interval for a difference between two means and since sample 1 and sample 2 are under 30, will be

$x_{1}-x_{2}$ ± $t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$

where

x₁ and x₂ are sample means

t is t-score

$S_{p}$ is estimate of standard deviation

n₁ and n₂ are the sample numbers

The estimate of standard deviation is calculated as

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} }$

where

s₁ and s₂ are sample standard deviation of each sample

Degrees of freedom is:

$df=n_{1}+n_{2}-2$

df = 12 + 9 - 2

df = 19

Checking t-table, with 90% Confidence Interval and df = 19, t = 1.729.

The mean and standard deviation for 12 unlogged forest plots are 17.5 and 3.53, respectively.

The mean and standard deviation for 9 logged plots are 13.66 and 4.5, respectively.

Calculating estimate of standard deviaton:

$S_{p}=\sqrt{\frac{(12-1)(3.53)^{2}+(9-1)(4.5)^{2}}{12+9-2} }$

$S_{p}=\sqrt{\frac{299.07}{19} }$

$S_{p}=$ 15.74

The difference between means is

$x_{1}-x_{2}$ = 17.5 - 13.66 = 3.84

Calculating the interval:

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $1.729.15.74.\sqrt{\frac{1}{12} +\frac{1}{9} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{\frac{21}{108} }$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = $27.21\sqrt{0.194}$

$t.S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}} }$ = 12

Then, interval for the difference in mean is 3.84 ± 12, which means the interval is between:

lower limit: 3.84 - 12 = -8.16

upper limit: 3.84 + 12 = 15.84

The interval is from -8.16 to 15.84.

5 0

3 years ago