What are 3 equivalent ratios of 14:2

bob bought bill an item that cost 193. bill gave him 200. bob then took 100 out of the 200 to buy bill an item for 93. how much

jekas [21]

Answer:Bob Owes $186 to the store he purchased an item at

Step-by-step explanation:

5 0

3 years ago

Has anyone taken the math kangaroo competition

kirill [66]

No i haven’t i’m sorry,what even is that tho

6 0

3 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below.

garik1379 [7]

Answer:

The 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =9.2$ represent the sample mean 1

$\bar X_2 =8.8$ represent the sample mean 2

n1=27 represent the sample 1 size

n2=30 represent the sample 2 size

$\sigma_1 =0.3$ population standard deviation for sample 1

$\sigma_2 =0.1$ population standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =9.2-8.8=0.4$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $t_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$0.4-1.96\sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}=0.281$

$0.4+1.96\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=0.519$

So on this case the 95% confidence interval would be given by $0.281 \leq \mu_1 -\mu_2 \leq 0.529$

6 0

3 years ago

Y is directly related to X, and Yis 81 when X is 27.<br> The constant of variation is

Arisa [49]

Answer:

3

Step-by-step explanation:

A direct variation is of the form

y = kx

We know x and y so we can find k

81 = k *27

Divide each side by 27

81/27 = 27k/27

3=k

The constant of variation is 3

5 0

4 years ago

Set of data with 350,000 numbers is normally

Rudiy27

Step-by-step explanation:

given a normal distribution with the given parameters the probability (= the % of the area of the distribution curve) for a number to be between 203 and 1803 is

0.9987

so, 99.87% of all numbers are expected to be in that range.

for 350,000 numbers that means

350,000×0.9987 = 349,545 numbers are expected to be between 203 and 1803.

7 0

2 years ago