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4 years ago

Jake is applying to rent an apartment. Which information about Jake could appear on his credit report?

Mathematics

1 answer:

bulgar [2K]4 years ago

7 0

B, C, & D.
To my knowledge you do not have access to the country where someone was born.

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Y = 0.115x + 0.0045 , y = 0.499x – 0.22<br> solve for x

enyata [817]

Answer:

x=0.5612

Step-by-step explanation:

-Given the functions:

$y=0.115x+0.0045\ \ \ \ ...i\\\\y=0.499x-0.22\ \ \ \ \ ...ii$

We subtract equation i from ii to eliminate y and solve for x:

$ii-i\\\\0=0.384x-0.2155\\\\0.384x=0.2155\\\\x=0.5612$

Hence, the value of x is 0.5612

6 0

3 years ago

PLEASE HELP ME people I beg you

Paraphin [41]

Equation of circumference is C=2pi times radius
So radius being 32 and pi being 3.14
You would do
2x3.14x32
= 200.96
Hope I helped

3 0

4 years ago

2/9 x __ = 3/15. 2/5 x __ = 7/12.

vodka [1.7K]

Answer:

1 x= 9/10 or 0.9

2 x= 35/24 or 1 11/24

3 0

3 years ago

What is the answer <br> How do you solve this

Talja [164]

The term for this sequence is 2.

It's an easy pattern!..

The second term is two times two.

The third term is two times three and so on..

4 0

4 years ago

Laboratory experiment shows that the life of the average butterfly is normally distributed with a mean of 18.8 days and a standa

inessss [21]

Answer:

a) 0.4164

Step-by-step explanation:

Mean lifespan (μ) = 18.8 days

Standard deviation (σ) = 2 days

For any given lifespan 'X', the z-score is:

$z=\frac{X- \mu}{\sigma} \\z=\frac{X- 18.8}{2}$

For X=12.04 days:

$z=\frac{12.04 - 18.8}{2} \\z=-3.38$

A z-score of -3.38 falls in the 0.036-th percentile of a normal distribution.

For X=18.38 days:

$z=\frac{18.38 - 18.8}{2} \\z=-0.21$

A z-score of -3.38 falls in the 41.68th percentile of a normal distribution.

The probability that a butterfly lives between 12.04 and 18.38 days is

$P=41.68-0.036\\P=41.64\%\ or\ 0.4164$

6 0

3 years ago