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4 years ago

What is 15 square root to the nearest tenth

Mathematics

1 answer:

Alex787 [66]4 years ago

8 0

Mathematically the answer would be a never ending decimal (3.87298...) but in this case it would be 3.9

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A bit is a 0 or a 1. A bit string of length 8 is a sequence of 8 digits, all of which are either 0 and 1.

andrezito [222]

Answer:

a) N = 2^8 = 256

b)N = 2^8 + 2^7 +2^6 +2^5 +2^4 +2^3 +2^2 +2^1

Step-by-step explanation:

Given;

String of length 8, with two options (0 or 1) each.

a) for the number of bits of strings of length 8.

Each digit of the 8 digit string has 2 options.

For 8 digit, we have

N = 2×2×2.... = 2^8 =256

b) for bit of strings of length 8 or less.

For n string = 2^n

For n < 8

The number of bit strings for length 8 or less are;

N = N8 +N7 + N6 + ... +N1

N = 2^8 + 2^7 +2^6 +2^5 +2^4 +2^3 +2^2 +2^1

8 0

3 years ago

Which best represent the solution for the inequality shown? x−2−3≤5

Murljashka [212]

Answer:

x≤10

Step-by-step explanation:

7 0

3 years ago

Suppose μ1 and μ2 are true mean stopping distances at 50 mph for cars of a certain type equipped with two different types of bra

Stella [2.4K]

Answer:

The 95% confidence interval would be given by $-23.44 \leq \mu_1 -\mu_2 \leq -8.16$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =114.1$ represent the sample mean 1

$\bar X_2 =129.9$ represent the sample mean 2

n1=5 represent the sample 1 size

n2=2 represent the sample 2 size

$s_1 =5.08$ sample standard deviation for sample 1

$s_2 =5.37$ sample standard deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =114.1-129.9=-15.8$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_1 +n_2 -1=5+5-2=8$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_{\alpha/2}=\pm 2.31$