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3 years ago

Need emergency help!!

Mathematics

1 answer:

Dafna1 [17]3 years ago

5 0

A. 18.99 b 16.676 c 30.11 d 13.801 can't answer the rest right now I'm getting ready for school

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Use a transformation to solve the equation. b + 0.6 = –1

Nina [5.8K]

You answer should be :

-1.6

Hope that helped!

8 0

3 years ago

(2y-2)÷x evaluate x=3 and y=4

Ratling [72]

Answer:

= 2

Step-by-step explanation:

Step 1- Substitute the variables

(2(4)-2)÷3

Step 2- Multiply

(2(4)-2)÷3

Step 3- Subtract

(8-2)÷3

Step 4- Divide

6÷3=2

7 0

3 years ago

Steve and Karla are studying the text messaging habits of middle school students. They collected data about the number of messag

Bess [88]

Answer:

The correct answer is A.

A) The IQR of Karla's data in 13.

Step-by-step explanation:

The interquartile range can be defined as the difference of upper quartile and lower quartile range. If we want to find the IQR from the box plot, we can simply see the length of the box from the box plot, as it represent the IQR.

We can clearly see that:

IQR of Steve's data is 45 - 31 = 14, hence the second statement is incorrect

IQR of Karla's data is 52 - 39 = 13, hence the first statement is CORRECT

The difference of medians is 43 - 36 = 7

Hence the difference of medians is not the half or twice of IQR ranges of both data sets

5 0

4 years ago

The manufacturer of an airport baggage scanning machine claims it can handle an average of 560 bags per hour. (a-1) At α = .05 i

ioda

Answer:

The calculated value t = 1.76 < 2.131 at 0.05 level of significance

Null hypothesis is accepted

The manufacturer’s claim is greater than 560 bags per hour

Step-by-step explanation:

Explanation:-

Given sample size 'n' =16

Given the manufacturer of an airport baggage scanning machine claims it can handle an average of 560 bags per hour.

mean of the Population 'μ' = 560

Mean of the sample Χ⁻ = 538

sample standard deviation' S' = 50

Null hypothesis:H₀:μ > 560

Alternative Hypothesis:H₁ : :μ < 560 (left tailed test)

Test statistic

$t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }$

$t = \frac{538-560 }{\frac{50}{\sqrt{16} } } = -1.76$

|t| = |-1.76| = 1.76

Degrees of freedom

γ = n-1 =16-1 =15

$t_{\frac{\alpha }{2} } = t_{\frac{0.05}{2} } =t_{0.025} =2.131$

Conclusion:-

The calculated value t = 1.76 < 2.131 at 0.05 level of significance

Null hypothesis is accepted

The manufacturer’s claim is greater than 560 bags per hour

6 0

4 years ago

Please help me i’ll mark brainliest, comment, and rate 5 stars

leonid [27]

Answer:

c...........................

5 0

3 years ago