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3 years ago

Express x2-8x+5 in the form (x-a)2 -b where a and b are intergers

Mathematics

1 answer:

Soloha48 [4]3 years ago

3 0

(x-4)^2-11

How?

$\\ \sf\longmapsto (x-4)^2-11$

$\\ \sf\longmapsto x^2-8x+16-11$

$\\ \sf\longmapsto x^2-8x+5$

Verified

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A study showed that 69% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate

Valentin [98]

Answer:

We conclude that supermarket ketchup was not as good as the national brand ketchup.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 100

p = 69% = 0.69

Alpha, α = 0.05

Number of stating that the supermarket brand was as good as the national brand , x = 56

a) First, we design the null and the alternate hypothesis

$H_{0}: p = 0.69\\H_A: p \neq 0.69$

This is a two-tailed test.

b) Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{56}{100} = 0.56$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.56-0.69}{\sqrt{\frac{0.69(1-0.69)}{100}}} = -2.810$

Now, we calculate the p-value from the table.

P-value = 0.0049

c) Since the p-value is lower than the significance level, we fail to accept the null and reject it.

Thus, we conclude that supermarket ketchup was not as good as the national brand ketchup.

d) It need to be tested further whether the supermarket brand was worse than the national brand or better than the national brand.

6 0

3 years ago

Jesse ate 100 donuts in five days. Each day 6 more donuts than the day before. How many donuts did she eat each day?

rusak2 [61]

Answer:

She ate 8 donuts the first day, 14 on the second, 20 on the third, 26 on the fourth, and 32 on the fifth day

Step-by-step explanation:

Over the course of 5 days, she ate 100 donuts, each day eating 6 more than the day before. Let n be the number of donuts she ate on the first day, then she ate...

Day 1: n donuts

Day 2: n + 6 donuts

Day 3: n + 12 donuts

Day 4: n + 18 donuts

Day 5: n + 24 donuts

Add the days together and get the equation...

n + (n + 6) + (n + 12) + (n + 18) + (n + 24) = 100

Now combine like terms and solve for n...

5n + 60 = 100

5n = 40

n = 8

She ate 8 donuts the first day, 14 on the second, 20 on the third, 26 on the fourth, and 32 on the fifth day

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3 years ago

For every $5 Micah saves, his parents give him $10. What is the constant of proportionality

liberstina [14]

When he gets five he gets 2 mor fives

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3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

6 0

3 years ago

Please help answer ASAP

Anarel [89]

8.95 times 1.20 = 10.74

10.74 + 4.31 = $15.05

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3 years ago