Question below here you can rear if not aks me please and thank you

Which number is NOT in the solution set of the inequality x > −15?

Anettt [7]

Answer: D).-20

Step-by-step explanation:

5 0

3 years ago

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Mary takes 9 grapes from rohin and then decides to give 4 back draw a model for the subtraction problem using integer chips

kaheart [24]

That should be correct

3 0

3 years ago

Convert 6 kilometers to yards

NISA [10]

1 km=1093.61yards

6 km= 6*1093.61yards
6 km= 6561.66yards

Answer= 6561.66 yards

7 0

3 years ago

Hassan invested $900 in an account that pays 4.75% interest compounded annually. Assuming no deposits or withdrawals are made, f

mestny [16]

$~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$900\\ r=rate\to 4.75\%\to \frac{4.75}{100}\dotfill &0.0475\\ n= \begin{array}{llll} \textit{times it compounds per year} \end{array}\dotfill &1\\ t=years\dotfill &9 \end{cases} \\\\\\ A=900\left(1+\frac{0.0475}{1}\right)^{1\cdot 9}\implies A=900(1.0475)^9\implies A\approx 1366.6$

6 0

2 years ago

Read 2 more answers

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

<u>e)</u> $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

2 years ago

Question below here you can rear if not aks me please and thank you ​

Question below here you can rear if not aks me please and thank you