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salantis [7]
2 years ago
8

Fake Question So It Doesnt Get Taken Down, What is 40992981x9977, Real Question, How do I change my pfp?

Mathematics
2 answers:
DedPeter [7]2 years ago
7 0

go to where the bell is and where it says "ask question" and then do you see your little icon thingy? right there, click it.  hit edit profile. and then

it shoulld say this

Edit your profile

Preferences

Send info to email

Password

Your email

Profile picture

hit profile picture and boom it says upload and all that

Ket [755]2 years ago
5 0

Answer:

go to Edit your profile

Step-by-step explanation:

You might be interested in
Solve for m <br>5m - 10m = 20<br><br>can anyone help me out?​
sweet-ann [11.9K]

Answer:

m=-4

Step-by-step explanation:

5m-10m=-10m+5m=-(10m-5m)=-5m=20

divide by 5 on both sides, -m=4

miltiply by -1 on both sides, m=-4

8 0
2 years ago
Read 2 more answers
Jana And Jordan had 23 oranges Jana had 7 fewer oranges than Jordan how many oranges did each girl have
Solnce55 [7]

Answer:

Jana had 8 oranges, and Jordan had 15 oranges.

Step-by-step explanation:

First, identify what you know:

1) In total, Jana and Jordan had 23 oranges.

2) Jana had 7 less oranges than Jordan.

We can create a formula, where x equals the number of oranges Jana has.

23 = x + (x + 7)

16 = x + x (subtracted 7 from both sides)

x = 8 (divided both sides by 2)

So, now we know Jana had 8 oranges, which is 7 less than Jordan.

8 + 7 = ?

? = 15

Jana had 8 oranges, while Jordan had 15, for a total of 23 oranges.

3 0
3 years ago
Simplify the following:
aivan3 [116]

Answer:

A

Step-by-step explanation:

6 0
3 years ago
A force of $20$ newtons will stretch a spring $5$ centimeters. What is the total number of centimeters that a force of $60$ newt
Aleksandr-060686 [28]

Answer:

Step-by-step explanation:

Hooke's Law applies here, which is a linear relationship. It would be easier to solve this using proportions with N of force on the top and the amount of stretch in cm on the bottom. Set up with our unknown, the number of cm:

\frac{N}{cm}:\frac{20}{5}=\frac{60}{x} and cross multiply:

20x = 300 so

x = 15 cm

6 0
2 years ago
Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams
Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0
3 years ago
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