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bagirrra123 [75]
3 years ago
15

What’s .9466 to the nearest hundredth I asked this before and someone gave me answer but it seems different then what I’ve learn

ed not sure if it’s typo but I wanted to check. Real quick before making it offical
Mathematics
2 answers:
Tju [1.3M]3 years ago
6 0

.95

5 and over (whether it's a decimal or not) is rounded up to the next number.

Please mark brainliest if this was helpful.

svetlana [45]3 years ago
4 0
The hundredth place is two places away from the decimal, so the answer would be 0.95.
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mixas84 [53]

Answer:

5

Step-by-step explanation:

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3 years ago
Between the years of 1947 and 1956, earthenware jars containing what are known as the Dead Sea Scrolls were found in caves along
lidiya [134]

Answer:

They are right, the time calculated using the exponential decay equation is 1948.6 years.

Step-by-step explanation:

The time can be calculated using the exponential decay equation:

N_{(t)} = N_{0}e^{-\lambda t}     (1)

<u>Where</u>:

N(t): is the quantity of C-14 at time t

N₀: is the initial quantity of C-14

λ: is the decay constant  

The decay constant is:

\lambda = \frac{ln(2)}{t_{1/2}}   (2)    

By entering equation (2) into (1) and solving for t, we have:

t = \frac{t_{1/2}*ln(N_{t}/N_{0})}{ln(2)} = \frac{5730*ln(0.79)}{ln(2)} = 1948.6 y

Therefore, the time of the scrolls estimated by the archeologists is right, since the time calculated using the exponential decay equation is 1948.6 years.      

I hope it helps you!

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3 years ago
A semi-truck is carrying 555 bags of chips. If each bag of chips has 321 chips, how many chips are in the semi-truck?
klemol [59]
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3 years ago
An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 37 type K batteries and a sample of 58
Olegator [25]

Answer:

Hypothesis Test states that we will accept null hypothesis.

Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, \mu_1 = true mean voltage for type K batteries.

           \mu_2 = true mean voltage for type Q batteries.

So, Null Hypothesis, H_0 :  \mu_1 = \mu_2 {mean voltage for these two types of

                                                        batteries is same}

Alternate Hypothesis, H_1 : \mu_1 \neq \mu_2 {mean voltage for these two types of

                                                          batteries is different]

<em>The test statistics we use here will be :</em>

                     \frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  follows t_n__1 + n_2  -2

where, X_1bar = 8.54         and     X_2bar = 8.69

                s_1  = 0.225       and         s_2     =  0.725

               n_1   =  37           and         n_2     =  58

               s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} } =   \sqrt{\frac{(37-1)0.225^{2}+(58-1)0.725^{2}  }{37+58-2} }  =  0.585               Here, we use t test statistics because we know nothing about population standard deviations.

     Test statistics =  \frac{(8.54-8.69) - 0 }{0.585\sqrt{\frac{1}{37}+\frac{1}{58}  } } follows t_9_3

                             = -1.219

<em>At 0.1 or 10% level of significance t table gives a critical value between (-1.671,-1.658) to (1.671,1.658) at 93 degree of freedom. Since our test statistics is more than the critical table value of t as -1.219 > (-1.671,-1.658) so we have insufficient evidence to reject null hypothesis.</em>

Therefore, we conclude that mean voltage for these two types of batteries is same.

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ziro4ka [17]
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