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lara31 [8.8K]
3 years ago
10

Please help me with this ​

Mathematics
2 answers:
QveST [7]3 years ago
5 0

Answer:

Dimensions are 10 x 9 x 7 and 9 x 4 x 5

Step-by-step explanation:

Divide the prism on the line that is 12cm

nasty-shy [4]3 years ago
3 0
The top prism is 9 x 4 x 5 = 180
the bottom prism is 10 x 9 x 7 = 630
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GalinKa [24]
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0.15625  \times 64
5 0
3 years ago
Read 2 more answers
Solve the system by using a matrix equation (Picture provided)
Katyanochek1 [597]

Answer:

Option b is correct (8,13).

Step-by-step explanation:

7x - 4y = 4

10x - 6y =2

it can be represented in matrix form as\left[\begin{array}{cc}7&-4\\10&-6\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}4\\2\end{array}\right]

A= \left[\begin{array}{cc}7&-4\\10&-6\end{array}\right]

X= \left[\begin{array}{c}x\\y\end{array}\right]

B= \left[\begin{array}{c}4\\2\end{array}\right]

i.e, AX=B

or X= A⁻¹ B

A⁻¹ = 1/|A| * Adj A

determinant of A = |A|= (7*-6) - (-4*10)

                                    = (-42)-(-40)

                                    = (-42) + 40 = -2

so, |A| = -2

Adj A=  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]

A⁻¹ =  \left[\begin{array}{cc}-6&4\\-10&7\end{array}\right]/ -2

A⁻¹ =  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right]

X= A⁻¹ B

X=  \left[\begin{array}{cc}3&-2\\5&-7/2\end{array}\right] *\left[\begin{array}{c}4\\2\end{array}\right]

X= \left[\begin{array}{c}(3*4) + (-2*2)\\(5*4) + (-7/2*2)\end{array}\right]

X= \left[\begin{array}{c}12-4\\20-7\end{array}\right]

X= \left[\begin{array}{c}8\\13\end{array}\right]

x= 8, y= 13

solution set= (8,13).

Option b is correct.

3 0
3 years ago
Express the location of the point on the number line as both a fraction and a decimal.
scoundrel [369]

Answer:

Fraction 1/6. Decimal .6 or 1.6

Step-by-step explanation:

3 0
3 years ago
Chloe made 8 out of 20 baskets in her basketball game. What is her<br> experimental probability? ???
Lerok [7]

10 would be 50% so 1 is 5% 5x8 =40

Her experimental probability is 40%

8 0
3 years ago
The average daily volume of a computer stock in 2011 was million​ shares, according to a reliable source. A stock analyst believ
igomit [66]

Complete question :

The average daily volume of a computer stock in 2011 was p = 35.1 million shares, according to a reliable source. A stock analyst believes that the stock volume in 2014 is different from the 2011 level. Based on a random sample of 40 trading days in 2014, he finds the sample mean to be 30.9 million shares, with a standard deviation of s = 11.8 million shares. Test the hypotheses by constructing a 95% confidence interval. Complete parts (a) through (c) below. State the hypotheses for the test. Construct a 95% confidence interval about the sample mean of stocks traded in 2014.

Answer:

H0 : μ = 35.1 ;

H1 : μ < 35.1 ;

(26.488 ; 35.312)

Step-by-step explanation:

The hypothesis :

H0 : μ = 35.1

H1 : μ < 35.1

The confidence interval :

Xbar ± Margin of error

Xbar = 30.9

Margin of Error = Zcritical * s/sqrt(n)

Zcritical at 95% = 1.96

Margin of Error = 1.96 * (11.8/sqrt(40))

Margin of Error = 4.412

Lower boundary :

30.9 - 4.412 = 26.488

Upper boundary :

30.9 + 4.412 = 35.312

Confidence interval = (26.488 ; 35.312)

Since the population mean value exists within the interval, the we fail to reject the Null.

5 0
3 years ago
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