If the total cost C for p pounds of pepperoni if each pound cost 4.86

Help with algebraic equation and to solve it plz

lutik1710 [3]

4) fifteen less than y is 32

15 - y= 32

15 - (-17)= 32

two negatives become positive so 15+17= 32

7 0

3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

2 years ago

Use the following image to assist in your answer.<br> a =<br> - 6<br> - 9<br> - 4

vfiekz [6]

Step-by-step explanation:

Pythagoras' theorem for the smallest one :

$c^2 = 4^2 + 6^2$

$c^2 = 16 + 36$

$c^{2}$ = 52

Pythagoras' theorem for the middle one :

$b^{2}$ = $6^{2}$ + $a^{2}$

Pythagoras' theorem for the biggest one :

$(4+a)^2 = c^2 + b^2$

$16 + 8a + a^2 = 52 + b^2$

Using the formula before (for $b^2$ ) it becomes :

$16 + 8a + a^2 = 52 + (6^2 + a^2)$

$16 + 8a + a^2 = 52 + 6^2 + a^2$

$16 + 8a = 52 + 6^2$

16 + 8a = 52 + 36

16+8a = 88

8a = 88-16

8a = 72

$a = \frac{72}{8}$

a = 9

Verifying :

$b^2 = 6^2 + a^2$

$b^2 = 36 + 81$

$b^2 = 117$

$b^{2}$ = 117

The biggest one :

$(4+a)^2 = c^2 + b^2$

$(4+9)^2 = 52 + 117$

$13^2 = 169$

True

8 0

2 years ago

Frank kept track of the amount of money he earned each day for 2 weeks.the amounts,in dollars,are listed below.

Lena [83]

Answer:

Step-by-step explanation:

5 0

2 years ago

ASAP What is the lateral area of this regular octagonal pyramid?? Explain your answer if you can plz thanks

Archy [21]

The lateral area would be 298.7 cm².

The lateral area is the area of all of the lateral faces of the pyramid. There are 8 triangles making up the lateral faces. Each has a base of 6.6. The formula for the area of a triangle is

A=1/2bh,

so we still need the height of the triangle.

The height of each lateral triangle is the slant height of the pyramid. The slant height of the pyramid forms a right triangle with the height of the pyramid and the "radius" as it were of the pyramid. Thus we use the Pythagorean theorem:

8²+8²=h²
64+64=h²
128=h²
√128=√(h²)
8√2 = h

Substituting this into our area formula we have:
A=1/2(6.6)(8√2)

We will go ahead and multiply this by 8, since there are 8 lateral faces:
LA=8(1/2)(6.6)(8√2)
LA = 298.7

3 0

3 years ago

Read 2 more answers