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3 years ago

What’s the number for 2+1=

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

8 0

2+2 is 4 minus one that's 3 quick maths but 2+1 is equal to 3

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I want to earn $42,000 for a new car, I have 12,678, How much more money do I need to buy the car?

laila [671]

So, you have $12,678. However, you want to earn $42,000 for a new car. If you want to figure out how much money you need, you need to subtract the number of money you want to earn by the number of money that you already have:

$42,000 - 12, 678 = 29,322$

You would need $29,322 more in order to buy the car.

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3 years ago

Pls tell me it’s a,b,c,or d.Really need it thx

Serggg [28]

Answer: B and C are both correct so idk

Step-by-step explanation:

when you multiply the numbers outside of the choices, both answers in B and C are correct.

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3 years ago

What number best represents the temperature in Anchorage, Alaska of below 12 degrees?

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3 years ago

Write the following decimals in order from greatest to least.

Colt1911 [192]

Answer:

0.6, 0.06, 0.04, 0.004

Step-by-step explanation:

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3 years ago

Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or alwa

andrew-mc [135]

Answer:

We are given that 81% of respondents said that their jobs were sometimes or always stressful.

So, p = 0.81

(a) Approximate the probability that 150 or fewer workers find their jobs stressful.

x = 150

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{150}{175} =\widehat{p}$

$0.8571=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8571-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=1.5882$

Refer the z table

So, P(z<150)=0.9429

So, the probability that 150 or fewer workers find their jobs stressful is 0.9429

b) Approximate the probability that more than 140 workers find their jobs stressful.

x = 140

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{140}{175} =\widehat{p}$

$0.8=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=-0.3372$

P(z<140) = 0.3707

P(z>140)=1-P(z<140)=1-0.3707=0.6293

So, the probability that more than 140 workers find their jobs stressful is 0.6293

c) Approximate the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive.

x = 146

p = 0.81

n = 175

So, $\frac{x}{n} =\widehat{p}$

$\frac{146}{175} =\widehat{p}$

$0.8342=\widehat{p}$

Formula : $z=\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$z=\frac{0.8342-0.81}{\sqrt{\frac{0.81(1-0.81)}{175}}}$

$z=0.816$

From part a) P(z<150)=0.9429

So,P(z<150)-P(z<146)=0.9429 - 0.8342=0.1087

Hence the probability that the number of workers who find their jobs stressful is between 146 and 150 inclusive is 0.1087

7 0

3 years ago