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3 years ago

Can someone please solve this for me

Mathematics

2 answers:

Savatey [412]3 years ago

5 0

Answer:

m<JKD = 22 degrees

Step-by-step explanation:

21x + 5 + 4x + 2 = 132

25x + 7 = 132

25x = 125

x = 5

m<JKD = 4x + 2

4(5) + 2

20 + 2

ICE Princess25 [194]3 years ago

4 0

Ghhoggliidtttuyryuduyuyiwuuvvkh I just wanted to let

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SCORPION-xisa [38]

Answer:

the correct option is choice D

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2 years ago

Rachel Rose 3.2 miles each weekday and 1.5 miles each day of weekend. how many miles will she have run in 6 weeks.

faltersainse [42]

Easiest way is to find out first how many she runs in a week. For that you can multiply the number of hours she runs a weekday by 5. (3.2*5) That should be 16. Now, since you want to know how many miles she will run in a span of 6 weeks you must multiply it by 6 now. That is 96. Since you know that, let's move on to the weekends. It is 1.5 per weekend day or 3 per weekend. Now you have to multiply 3 by 6 because you want to know for six weeks. Since you have both your numbers now, 18 and 96, you can add them to make a final of 114.

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3 years ago

A pallet requires 2 mins 15 seconds to prepare for shipment and 30 seconds to load on a trailer how long will it take to prepare

Reika [66]

21.4333333333 thats wut i got hope i helped :-)

6 0

3 years ago

Read 2 more answers

5/6(z+10)=20 what is z plzz I WILL GIVE YOU 16PTS

aleksandrvk [35]

Answer:

1/6

Step-by-step explanation:

3 0

3 years ago

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

substitute values and we get S² = 40.988

$t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$

t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

6 0

3 years ago