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jenyasd209 [6]
3 years ago
5

The distance between itahari and Damak is

Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

8 km by motorbike.

Step-by-step explanation:

The distance between itahari and Damak is  42 km.

Mr. Dhamala travelled 18.325 km by a  bus / 15.675 km by a taxi and the remaining distance by a motorbike.

Let he covered x distance by motorbike. So,

Remaining distance = Total distance -(distance covered by bus + distance by taxi)

= 42 -(18.325 +15.675 )

= 42-34

= 8 km

So, he travelled 8 km by motorbike.

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Mr Taylor brought back 180 Euros to the UK. The exchange rate was €1.2 to £1. How much did he get in £ sterling?
Firdavs [7]
You need to divide 180 by 1.2 
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makela plants 3/4 of her herb garden with basil and 1/4 with sage. explain why the fraction strips show that the whole garden is
KonstantinChe [14]
The garden has four quarters. three quarters are planted with basil. one quarter is planted with sage. The strips represent quarters. So that means that three strips represent three quarters. Well one strip represents one quarter.
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4 years ago
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Solve the matrix equation for a, b, c, and d. [1 2] [a b] [6 5][3 4] [c d]= [19 8]
Anit [1.1K]

Answer:

The answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

Step-by-step explanation:

\bold{\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]}

Solve the L.H.S part:

\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\\\\\\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]

After calculating the L.H.S part compare the value with R.H.S:

\left[\begin{array}{cc}a+2c&b+2d\\3a+4c&3b+4d\end{array}\right]= \left[\begin{array}{cc}6&5\\ 19&8\end{array}\right]} \\\\

\to a+2c =6....(i)\\\\\to b+2d =5....(ii)\\\\\to 3a+4c =19....(iii)\\\\\to 3b+4d = 8 ....(iv)\\\\

In equation (i) multiply by 3 and subtract by equation (iii):

\to 3a+6c=18\\\to 3a+4c=19\\\\\text{subtract}... \\\\\to 2c = -1\\\\\to  c= - \frac{1}{2}

put the value of c in equation (i):

\to a+ 2 (- \frac{1}{2})=6\\\\\to a- 2 \times \frac{1}{2}=6\\\\\to a- 1=6\\\\\to a =6 +1\\\\\to a = 7\\

In equation (ii) multiply by 3 then subtract by equation (iv):

\to 3b+6d=15\\\to 3b+4d=8\\\\\text{subtract...}\\\\\to 2d = 7\\\\\to d= \frac{7}{2}\\

put the value of d in equation (iv):

\to 3b+4 (\frac{7}{2})=8\\\\\to 3b+4 \times \frac{7}{2}=8\\\\\to 3b+14=8\\\\\to 3b =8-14\\\\\to 3b = -6\\\\\to b= \frac{-6}{3}\\\\\to b= -2

The final answer is "\bold{\left[\begin{array}{cc}a&b\\c&d\end{array}\right] = \left[\begin{array}{cc}7&-2\\ -\frac{1}{2}&\frac{7}{2}\end{array}\right]}".

4 0
3 years ago
An independent-measures research study was used to compare two treatment conditions with n= 12 participants in each treatment. T
Maurinko [17]

Answer:

(a) The data indicate a significant difference between the two treatments.

(b) The data do not indicate a significant difference between the two treatments.

(c) The data indicate a significant difference between the two treatments.

Step-by-step explanation:

Null hypothesis: There is no difference between the two treatments.

Alternate hypothesis: There is a significant difference between the two treatments.

Data given:

M1 = 55

M2 = 52

s1^2 = 8

s2^2 = 4

n1 = 12

n2 = 12

Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(12-1)8 + (12-1)4] ÷ (12+12-2) = 132 ÷ 22 = 6

Test statistic (t) = (M1 - M2) ÷ sqrt [pooled variance (1/n1 + 1/n2)] = (55 - 52) ÷ sqrt[6(1/6 + 1/6)] = 3 ÷ 1.414 = 2.122

Degree of freedom = n1+n2-2 = 12+12-2 = 22

(a) For a two-tailed test with a 0.05 (5%) significance level and 23 degrees of freedom, the critical values are -2.069 and 2.069.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 falls outside the region bounded by the critical values.

(b) For a two-tailed test with a 0.01 (1%) significance level and 23 degrees of freedom, the critical values are -2.807 and 2.807.

Conclusion:

Fail to reject the null hypothesis because the test statistic 2.122 falls within the region bounded by the critical values.

(c) For a one-tailed test with 0.05 (5%) significance level and 23 degrees of freedom, the critical value is 1.714.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 is greater than the critical value 1.714.

6 0
3 years ago
A koala bear eats at a steady rate 15 leaves in 2 hours. In how many minutes does the koala eat 10 leaves?
ZanzabumX [31]

Answer:

80 minutes

Step-by-step explanation:

1 hour=60 minutes

2 hours=120 minutes

-------------------------------

15/120=10/x

cross product

120*10=15*x

1200=15x

x=1200/15

x=80

8 0
3 years ago
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