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3 years ago

The product of -2 and a number minus six is greater than -18

1 answer:

7 0

So, its a multiplying question - the product. Also there are multiple answers from x = negative anything to positive 14, so i chose a number.

The product of -2 and a number minus six is greater than -18.

-2 · (x - 6) > -18

-2 · (7 - 6) > -18

-2 · (1) > -18

-2 > -18

x = 7, or anything lower than 15.

On the number line btw, -2 is bigger than -18

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Maura is 5 years younger than her sister Cara. Seven years ago, Maura was half as old as her sister. How old is Maura now?

masha68 [24]

Let's call Maura's age now M and Cara's age now C.

We know Maura is five years younger than Cara. In symbols that is, M - 5 = C

We also know that 7 years ago Maura's age was half of Cara's. Maura's age 7 years ago was M-7 and Cara's age seven years ago was C-7. Since Maura's age (7 years ago) was half of Cara's we can write in symbols: $M-7= \frac{1}{2}(C-7)$

Let's take that last equation and substitute C - 5 for M (this is because according to our first equation these are equal). When we do this we get an equation with only C as the variable and solve for C as follows:

$M-7= \frac{1}{2}(C-7)$
$C-5-7= \frac{1}{2}(C-7)$
$C-12= \frac{1}{2}C-( \frac{1}{2})(7)$
$C-12= .5C-3.5$
$.5C-12= -3.5$
$.5C= 8.5$
$C= 17$

Cara is 17. Since Maura is 5 years younger she is 12.

As a check, seven years ago Cara was 10 and Maura was 5. It is the case that Maura was half Cara's age seven years ago.

4 0

3 years ago

Pls help it due ASAP <br> Show workings <br> Pleaseeeeeeeeeeeee

Talja [164]

D
5 + 29 = 34/2 = 17
7 - 15 = -8/2 = -4
(17,-4)

3 0

3 years ago

A ski resort claims that there is a 75% chance of snow on any given day in january and that snowfall happens independently from

creativ13 [48]

The mean exists at 3 and the standard deviation exists 0.87.

Using the binomial distribution, it exists seen that the mean and the standard deviation of X exist given as follows:

$$\mu_{X}=3, \sigma_{X}=0.87$

<h3>What is the binomial probability distribution?</h3>

The expected value of the binomial distribution exists:

$\mu_{X}=n p$$

The standard deviation of the binomial distribution exists:

$\sigma_{X}=\sqrt{n p(1-p)}$$

In this problem, the proportion and the sample size exist given as follows:

p = 0.75

n = 4

Therefore, the mean and the standard deviation exist given by:

$$&\mu_{X}=n p=4(0.75)=3 \\$

$$&\sigma_{X}=\sqrt{n p(1-p)}$

$$=\sqrt{4(0.75)(0.25)}=0.87$

The mean exists at 3 and the standard deviation exists 0.87.

To learn about the binomial distribution refer to:

brainly.com/question/24863377

#SPJ4

7 0

2 years ago

Read 2 more answers

There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on

ELEN [110]

Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools is =32±2.7280

i.e. C.I. [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

True mean =32

Standard deviation=11

And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

C.I.=mean±Z[Standard deviation/Sqrt(n)]

=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

=32±2.7280

i.e. C.I. [29.3,34.7]

Now Here to commit on calculated interval we should check value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

Z-score=(sample mean -true mean)/[standard deviation/Sqrt(n)]

=(34.86-32)/[11/Sqrt(44)]

Z=1.72465

For ,

P(Z≥1.72465)=0.08544

For 0.05 significance level the results are not significant at p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

Hence the results are not significant.

6 0

3 years ago

Find AD if BD=6 and DC=8

Ludmilka [50]

What is the figure
please give full question

4 0

3 years ago

Read 2 more answers