Find the volume: in cm 7cm high 3cm wide 2 cm long

Express the tangent of ∠U as a ratio of the given side lengths.

Kay [80]

The answer should be a

6 0

3 years ago

A rectangular garden has length and width as given by the expressions below. Length: 4 - 7 (3x + 4y) Width: 3x(-2y) Write a simp

Nezavi [6.7K]

Answer:

⇒ 8 − ⇒ 42x − ⇒ 56y − ⇒ 12xy

Step-by-step explanation:

Im smart

3 0

3 years ago

Read 2 more answers

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts o

Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A $(100-\alpha )\%$ confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a $(100-\alpha )\%$ probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the t-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

$\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}$

Here,

$\bar x$ = sample mean

s = sample standard deviation

n = sample size

$t_{\alpha/2,(n-1)}$ = critical value.

The degrees of freedom for the critical value is, (n - 1) = 7 - 1 = 6.

The significance level is: $\alpha =1-Confidence\ level=1-0.99=0.01$

The critical value is:

$t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143$

**Use the t-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753$

$\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379$

The 99% confidence interval for μ is:

$x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)$

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

6 0

4 years ago

Please help me with this??

Lina20 [59]

This problem is simple the correct choice for this problem will be a

4 0

3 years ago

Find the probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if:a) a

lara31 [8.8K]

Answer:

A) 0.0009765625

B) 0.0060466176

C) 2.7756 x 10^(-17)

Step-by-step explanation:

A) This problem follows a binomial distribution. The number of successes among a fixed number of trials is; n = 10

If a 0 bit and 1 bit are equally likely, then the probability to select in 1 bit is; p = 1/2 = 0.5

Now the definition of binomial probability is given by;

P(K = x) = C(n, k)•p^(k)•(1 - p)^(n - k)

Now, we want the definition of this probability at k = 10.

Thus;

P(x = 10) = C(10,10)•0.5^(10)•(1 - 0.5)^(10 - 10)

P(x = 10) = 0.0009765625

B) here we are given that p = 0.6 while n remains 10 and k = 10

Thus;

P(x = 10) = C(10,10)•0.6^(10)•(1 - 0.6)^(10 - 10)

P(x=10) = 0.0060466176

C) we are given that;

P((x_i) = 1) = 1/(2^(i))

Where i = 1,2,3.....,n

Now, the probability for the different bits is independent, so we can use multiplication rule for independent events which gives;

P(x = 10) = P((x_1) = 1)•P((x_2) = 1)•P((x_3) = 1)••P((x_4) = 1)•P((x_5) = 1)•P((x_6) = 1)•P((x_7) = 1)•P((x_8) = 1)•P((x_9) = 1)•P((x_10) = 1)

This gives;

P(x = 10) = [1/(2^(1))]•[1/(2^(2))]•[1/(2^(3))]•[1/(2^(4))]....•[1/(2^(10))]

This gives;

P(x = 10) = [1/(2^(55))]

P(x = 10) = 2.7756 x 10^(-17)

3 0

3 years ago