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fomenos
3 years ago
11

A varies directly as b. If A = 3 when b = 24, find b when A = 10

Mathematics
1 answer:
Lelechka [254]3 years ago
4 0
240................................ It had to be 20 characters so sorry for the large answer <span />
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Can someone help me please I’ll mark u as brainliest
Sunny_sXe [5.5K]
Hi yes what do u need help with??
6 0
3 years ago
A student needed to prepare 500mL of 1X TAE buffer to run a QC gel. The stock solution in the lab is 5X TAE. What volumes of sto
Xelga [282]

Answer:

you need 100ml of 5X TAE and 400ml of water.

Step-by-step explanation:

You need to use a rule of three:

C_1V_1=C_2V_2

where:

\left \{ {{C_1= 5X} \atop {C_2=1X}} \right.

and

\left \{ {{V_1 = V_{TAE}} \atop {V_2=500ml}} \right.

Therefore:

V_{TAE} = \frac{V_2*C_2}{C_1}

V_{TAE} = 100ml

Then just rest the TAE volume to the final Volume and you get the amount of water that you need to reduce the concentration.

3 0
3 years ago
Read 2 more answers
2/10times6/10. Please help
babunello [35]

Answer:

I believe the answer is 1 2/10

Hope this helps!



8 0
4 years ago
Read 2 more answers
Two similar solid shapes a and b are shown, the solids are made from the same matierial the surface area of b is 40.32cm^2 and a
Kisachek [45]

Answer:

4000g

Step-by-step explanation:

7 0
3 years ago
A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p
Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, H_0 : p \leq 0.29  {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, H_1 : p > 0.29  {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

          T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = proportion of chips that do not fail in the first 1000 hours of their use = 32%

            n = sample of chips = 1500

So, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.</em>

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0
4 years ago
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