All categories

2 years ago

Help please, I'm stuck

Mathematics

2 answers:

mrs_skeptik [129]2 years ago

7 0

Answer:

Z<-3 is the anwser to this question, good luck !

Vikki [24]2 years ago

3 0

Multiply through by LCM, 4

$\frac{ - 3z - 1}{2} \times 4 > \: \: \: \frac{4}{1} \times 4$

$2( - 3z - 1) > 16$

$- 6z - 2 > 16$

$- 6z > 16 + 2$

$- 6z > 18$

$\frac{ - 6z}{ - 6} < \frac{18}{ - 6}$

$z < - 3$

You might be interested in

Assume that you plan to use a significance level of α = 0.05 to test the claim that p1 = p2, use the given sample sizes and numb

11Alexandr11 [23.1K]

Since the data is insufficient. Let us denote that n1 = 100, n2 = 100X1 = 38. X2 = 40
So looking for the p value:
p = (38+ 40) / (100 + 100) = 0.39
z = (38/100 - 40/100)/√ (0.39*(1-0.39)*(1/100 + 1/100))= -0.2899
P-value = P (|z| > 0.2899) = 0.7718

7 0

3 years ago

A radioactive substance decreases in the amount of grams by one-third each year. If the starting amount of the

katovenus [111]

Answer:

The sequence is geometric. The recursive formula is $a_{n}=2/3a_{n-1}$

Step-by-step explanation:

In order to solve this problem, you have to calculate the amount of the substance left after the end of each year to obtain a sequence and then you have to determine if the sequence is arithmetic or geometric.

The substance decreases by one-third each year, therefore:

After 1 year:

$1452-\frac{1}{3}(1452)$

Using 1452 as a common factor and solving the fraction:

$1452(1-\frac{1}{3})=1452(\frac{2}{3})=968$

You can notice that in general, after each year the amount of grams is the initial amount of the year multiplied by 2/3

After 2 years:

$968(\frac{2}{3})=\frac{1936}{3}$

After 3 years:

$\frac{1936}{3}(\frac{2}{3})=\frac{3872}{9}$

The sequence is:

1452,968,1936/3,3872/9....

In order to determine if the sequence is geometric, you have to calculate the ratio of two consecutive terms and see if the ratio is the same for all two consecutive terms. The ratio is obtained by dividing a term by the previous term.

The sequence is arithmetic if the difference of two consecutive terms is the same for all two consecutive terms.

-Calculating the ratio:

For the first and second terms:

968/1452=2/3

For the second and third terms:

1936/3 ÷ 968 = 2/3

In conclussion, the sequence is geometric because the ratio is common.

The recursive formula of a geometric sequence is given by:

$a_{n}=ra_{n-1}$

where an is the nth term, r is the common ratio and an-1 is the previous term.

In this case, r=2/3

7 0

3 years ago

Figure ABCD is a rhombus. What other term describes figure ABCD?

Maslowich

Paralellogram and quadrilateral and polygon

4 0

3 years ago

Read 2 more answers

The basketball team at Martin Luther King High School won 12 of its first 15 games. Then it won its next 3 games. The percent of

aliya0001 [1]

12/15 games have been won then they won 3 which makes it then 15/18 because the denominator is how many games they've played while the numerator is how many games they one.

$\frac{15}{18} = \frac{x}{100}$
Cross multiply both...
$1500=18x$
divide each side by 18 to make the x by itself
Which gives you x=83.333%

4 0

3 years ago

A 90 percent confidence interval for the slope of a regression line is determined to be (-0.181, 1.529). Which of the following

GenaCL600 [577]

Answer: a. The correlation coefficient of the data is positive.

Step-by-step explanation:

Estimated slope of sample regression line = $\dfrac{Upper \ limit + lower\ limit}{2}$

Here , confidence interval : (-0.181, 1.529)

Estimated slope of sample regression line = $\dfrac{-0.181+1.529}{2}$

$=\dfrac{1.348}{2}\\\\=0.674\ \ \ \ [\text{ positive}]$

⇒Correlation coefficient(r) must be positive, So a. is true.

But, d. and e. are wrong(0.674 ≠ 0 or 1.348).

We cannot check residuals or its sum from confidence interval of slope of a regression line, so b is wrong.

We cannot say that scatterplot is linear as we cannot determine it from interval, so c. is wrong

So, the correct option : a. The correlation coefficient of the data is positive.

7 0

3 years ago

Help please, I'm stuck ​

Help please, I'm stuck