BRAINLIEST!!!!! HELPPP ME YALLLL

Denise rolled a number cube several times. The results are shown.

strojnjashka [21]

4/156 so
2/78 so
1/39 so approx. 2.5641%

5 0

3 years ago

Read 2 more answers

Is the function linear or exponential f(x) =8(1.2)

777dan777 [17]

Linear is the correct answer because you don't see any exponents

7 0

3 years ago

Factor:

nekit [7.7K]

Answer:

a) $(7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)$

b) $(7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)$

c) $a = 0.7\cdot t^{9}$

d) There are two possible answers only considering real coefficients:

(i) $16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)$

(ii) $14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)$

e) $(a^{3}+4)\cdot (a^{3}-4)$

Step-by-step explanation:

Now we proceed to solve each algebraic equation:

a) Factor $49\cdot m^{12}-16$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$49\cdot m ^{12}-16$

$(7\cdot m^{6})^ 2-4^{2}$

$(7\cdot m^{6}+4)\cdot (7\cdot m^{6}-4)$

b) Factor $49\cdot a^{2}\cdot b^{16}-1$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$49\cdot a^{2}\cdot b^{16}-1$

$(7\cdot a\cdot b^{8})^{2}-1^{2}$

$(7\cdot a\cdot b^{8}+1)\cdot (7\cdot a\cdot b^{8}-1)$

c) Nicole is factoring $0.49\cdot t^{18}-25$ by using the rule $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . What will she use for the value of $a$ ?

From this rule we find that $a^{2} = 0.49\cdot t^{18}$ , that is:

$a^{2} = 0.49\cdot t^{18}$

$a^{2} = \frac{49}{100}\cdot t^{18}$

$a = \frac{7}{10}\cdot t^{9}$

$a = 0.7\cdot t^{9}$

d) Which binomial is a difference of squares?

A binomial is a difference of square if and only if $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . There are two possible answers only considering real coefficients:

(i) $16\cdot x^{2}-81 = (4\cdot x +9)\cdot (4\cdot x - 9)$

(ii) $14\cdot x^{2}-81 =(\sqrt{14}\cdot x+9)\cdot (\sqrt{14}\cdot x - 9)$

e) Factor $a^{6}-16$

This binomial is of the form $a^{2}-b^{2} = (a+b)\cdot (a-b)$ . In this case, we can rewrite and factor the equation below:

$a^{6}-16$

$(a^{3})^{2}-4^{2}$

$(a^{3}+4)\cdot (a^{3}-4)$

6 0

3 years ago

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 270 day

dusya [7]

Answer:

a) 281 days.

b) 255 days

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 270, \sigma = 8$

(a) What is the minimum pregnancy length that can be in the top 8% of pregnancy lengths?

100 - 8 = 92th percentile.

X when Z has a pvalue of 0.92. So X when Z = 1.405.

$Z = \frac{X - \mu}{\sigma}$

$1.405 = \frac{X - 270}{8}$

$X - 270 = 1.405*8$

$X = 281$

(b) What is the maximum pregnancy length that can be in the bottom 3% of pregnancy lengths?

3rd percentile.

X when Z has a pvalue of 0.03. So X when Z = -1.88

$Z = \frac{X - \mu}{\sigma}$

$-1.88 = \frac{X - 270}{8}$

$X - 270 = -1.88*8$

$X = 255$

8 0

4 years ago

Graph this problem please and thank you

nirvana33 [79]

I am not able to graph but I will help you out. All you have to do to solve this is add the rule to the original x and y values. If the rule is (x+9, y-2) and the point is (-2,3) all we’d to do is plug it in.

-2+9=7
3-2=1
So the new point for P would be (7, 1)

If you need anyone help, lemme know. If you really can’t figure it out. I will grab my computer and graph it for you.

6 0

3 years ago