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3 years ago

6

I need help finding this problem out please

1 answer:

Sergio [31]3 years ago

3 0

Answer: what’s the problem

Step-by-step explanation:

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Complete the ratio tables.<br> 10<br> 20<br> 25<br> 50<br> 60

Misha Larkins [42]

Answer:

what do u want me to do

Step-by-step explanation:

6 0

3 years ago

What is the percent decrease from 60 to 39?

IrinaK [193]

X - the percent decrease

$60-60x=39 \ \ \ |-60 \\
-60x=-21 \ \ \ |\div (-60) \\
x=\frac{-21}{-60} \\ \Downarrow \\ x=\frac{-21}{-60}=\frac{-21 \div (-3)}{-60 \div (-3)}=\frac{7}{20}=\frac{7 \times 5}{20 \times 5}=\frac{35}{100}=35\%$

It's a 35% decrease.

7 0

3 years ago

Read 2 more answers

Please answer correctly !!!!!!!!! Will mark brainliest answer !!!!!!!!!!!!

Lena [83]

Answer:

I think that C) and D)

Step-by-step explanation:

I think this because if you look at 3.5 secounds it came down and after 1.75 secounds the ball hit the ground. A) wouldn't be because it came down at 3.5 secounds not coming up. B) wouldn't be because the ball droped after 1.75 secounds not come up.

3 0

3 years ago

Find the error (-8x^2y^2)^2= -64x^4y^4

ExtremeBDS [4]

Answer:

$(-8)^2\neq-64$

Step-by-step explanation:

Using the distributive property to expand, we have

$(-8x^2y^2)^2=(-8)^2(x^2)^2(y^2)^2.$

Squaring each term, we have

$(-8x^2y^2)^2=64x^4y^4.$

The mistake was that $8^2=64,$ not $-64.$ Using the fact that $8^2=64,$ the right answer would be

$\boxed{64x^4y^4}$ .

3 0

2 years ago

Read 2 more answers

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

or

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

or

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

or

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

7 0

2 years ago