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natka813 [3]
2 years ago
13

Help me pleaseeeeeeeeee

Mathematics
1 answer:
crimeas [40]2 years ago
3 0

Answer:

x=10 and x=-9

Step-by-step explanation:

Factor:

(x-10)(x+9)=0

x-10=0

x=10

and

x+9=0

x=-9

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Luke started a weight-loss program. The first week, he lost X pounds. the second week he lost 3/2 pounds less then 3/2 times the
irga5000 [103]

Luke started a weight-loss program.

The first week, he lost X pounds.

the second week he lost 3/2 pounds less then 3/2 times the pounds he lost the first week. so, he lost 3/2 x - 3/2

The third week, he lost 1 more pound than 3/4 of the pounds he lost the first week. so, he lost 3/4 x + 1

So, over 3 weeks he lost :

x + (3/2 x - 3/2) + (3/4 x + 1) = 13/4 x - 1/2

=========================================================

Liam started the program when Luke did.

The first week, he lost 1 pound less than 3/2 times the pounds luke lost the first week. so, he lost (3/2 x - 1)

The second week, he lost 4 pounds less than 5/2 times the pounds luke lost the first week. so ,he lost (5/2 x - 4)

The third week, he lost 1/2 more than 5/4 times the pounds luke lost the first week. so, he lost, ( 5/4 x + 1/2)

The expression for Liam's weight loss over the three weeks is =

= (3/2 x - 1) + (5/2 x - 4) + ( 5/4 x + 1/2) = 21/4 x - 9/2

=============================================================

Assuming they both lost the same number of pounds over the three weeks

So, equating the number of pounds lost by each one of them

so,

13/4 x - 1/2 = 21/4 x - 9/2

solve for x

13/4 x - 21/4 x = -9/2 + 1/2

-2x = -4

divide both sides by -2

so,

x = -4/-2 = 2

so,

The number of pounds luke lost in the fist week is 2 pounds

=============================================================

4. The number of pounds luke and liam each lost over the three weeks is

= 13/4 * 2 - 1/2 = 6 pounds

7 0
1 year ago
The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes
devlian [24]

Answer:

(a) There is no evidence to support the claim that the two machines produce rods with different mean diameters.

P-value is 0.818

(b) 95% confidence interval for the difference in mean rod diameter is (-0.17, 0.27).

This interval shows that the difference in mean is between -0.17 and 0.27.

Step-by-step explanation:

(a) Null hypothesis: The two machines produce rods with the same mean diameter.

Alternate hypothesis: The two machines produce rods with different mean diameter.

Machine 1

mean = 8.73

variance = 0.35

n1 = 15

Machine 2

mean = 8.68

variance = 0.4

n2 = 17

pooled variance = [(15-1)0.35 + (17-1)0.4] ÷ (15+17-2) = 11.3 ÷ 30 = 0.38

Test statistic (t) = (8.73 - 8.68) ÷ sqrt[0.38(1/15 + 1/17)] = 0.05 ÷ 0.218 = 0.23

degree of freedom = n1+n2-2 = 15+17-2 = 30

Significance level = 0.05 = 5%

Critical values corresponding to 30 degrees of freedom and 5% significance level are -2.042 and 2.042.

Conclusion:

Fail to reject the null hypothesis because the test statistic 0.23 falls within the region bounded by the critical values -2.042 and 2.042.

There is no evidence to support the claim that the two machines produce rods with different mean diameters.

Cumulative area of test statistic is 0.5910

The test is a two-tailed test.

P-value = 2(1 - 0.5910) = 2×0.409 = 0.818

(b) Difference in mean = 8.73 - 8.68 = 0.05

pooled sd = sqrt(pooled variance) = sqrt(0.38) = 0.62

Critical value (t) = 2.042

E = t×pooled sd/√n1+n2 = 2.042×0.62/√15+17 = 0.22

Lower limit of difference in mean = 0.05 - 0.22 = -0.17

Upper limit of difference in mean = 0.05 + 0.22 = 0.27

95% confidence interval for the difference in mean rod diameter is between a lower limit of -0.17 and an upper limit of 0.27.

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-2^2=-4\\
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