Simplify (√3 − √7)^2

Luis put out

solniwko [45]

Answer:

Step-by-step explanation:

Find out how much pie was leftover from each pie:

Apple pie: 3/4, 4 - 3 = 1, 1/4

Blackberry: 1/2, 2 - 1 = 1, 1/2

Blueberry: 1/8, 8 - 1 = 7, 7/8

Cherry: 5/8, 8 - 5 = 3, 3/8

Peach: 3/4, 4 - 3 = 1, 1/4

There are 1/4, 1/2, 7/8, 3/8, and 1/4 pies left. The LCD if these numbers is 8, since 2*4 = 8, 4*2 = 8, and 8*1 = 8

1/4 = 2/8

1/2 = 4/8

7/8 = 7/8

3/8 = 3/8

1/4 = 2/8

2 + 4 + 7 + 3 + 2 = 18, so 18/8 pies, or 2 2/8, or 2 1/4

5 0

3 years ago

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Find the squares of (y+3)

emmasim [6.3K]

Answer:

Since we know that

(a+b)

2

=a

2

+2ab+b

2

(a−b)

2

=a

2

−2ab+b

2

Now,

(i)

(x+3y)

2

=x

2

+2×x×3y+(3y)

2

=x

2

+6xy+9y

2

(ii)

(2x−5y)

2

=(2x)

2

−2×2x×5y+(5y)

2

=4x

2

−20xy+25y

2

(iii)

(a+

5a

1

)

2

=a

2

+2×a×

2a

1

+(

2a

1

)

2

=a

2

+1+

4a

2

1

(iv)

(2a−1a)=(1a)

2

(1a)

2

=1a

2

Answer

8 0

2 years ago

Inez says you can plot 4/0 on a number line, but Jeremiah says you cannot. Who is correct? Explain.

anyanavicka [17]

Answer:

Jeremiah is correct

Step-by-step explanati

because any number divided by 0 is 0 and can be represented on a number line.

6 0

3 years ago

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NEED HELP, LIMITED TIME, HURRY

11111nata11111 [884]

Answer:

9

Step-by-step explanation:

here you go!

6 0

3 years ago

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J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.

melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean for mail-order sales = $74.50

$\bar X_2$ = sample mean for internet sales = $84.40

$s_1$ = sample standard deviation for mail-order purchases = $17.25

$s_2$ = sample standard deviation for internet purchases = $21.25

$n_1$ = sample of sales receipts for mail-order purchases = 16

$n_2$ = sample of sales receipts for internet purchases = 9

Also, $s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} }$ = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by ( $\mu_1-\mu_2$ ).

Now, 99% confidence interval for ( $\mu_1-\mu_2$ ) is given by;

= $(\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

= $(74.50-84.40) \pm (2.807 \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})$

= [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0

3 years ago