Write an expression that can be used to multiply 6 times 198 mentally

Which point on the number line represents the product (5)(-2)(-1)2

Thepotemich [5.8K]

The Right Answer:

We need to mutiply all the numbers.

5*-2 = -10

-10*-1 = 10

The Wrong Answer:

(-2)*(-1) gives +2 (a negative number multiplied for another negative gives a positive number).

2*5 = +10

5 0

1 year ago

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A liquid dietary product implies in its advertising that use of the product for one month results in an average weight loss of a

BigorU [14]

Answer:

Following are the responses to the given question:

Step-by-step explanation:

Please find the table in the attached file.

mean and standard deviation difference: $\bar{d}=\frac{\Sigma d}{n} =\frac{-4-6-.......-4-4}{8}=-4.125 \\\\S_d=\sqrt{\frac{\Sigma (d-\bar{d})^2 }{n-1}}=\sqrt{\frac{(-4 + 4.125)^2 +.......+(-4 +4.125)^2 }{8-1}}= 1.246$

For point a:

hypotheses are:

$H_0 : \mu_d \geq -3\\\\H_a : \mu_d < -3\\\\$

degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha = 0.05$ , reject null hypothesis if $t$ .

test statistic:

$t=\frac{\bar{d}-\mu_d }{\frac{s_d}{\sqrt{d}}}=\frac{ -4.125- (-3)}{\frac{1.246}{ \sqrt{8}}} =-2.55$

because the $t=-2.553$ , removing the null assumption. Data promotes a food product manufacturer's assertion with a likelihood of Type 1 error of 0.05.

For point b:

From t table, at $\alpha =0.01$ , removing the null hypothesis if $t$ .

because $t=-2.553 >-2.908$ , fail to removing the null hypothesis.

The data do not help the foodstuff producer's point with the likelihood of a .01-type mistake.

For point c:

Hypotheses are:

$H_0: \mu_d \geq -5\\\\H_a: \mu_d < -5$

Degree of freedom:

$df=n-1=8-1=7$

From t table, at $\alpha =0.05$ , removing the null hypothesis if $t$ .

test statistic: $t=\frac{\bar{d}-\mu_d}{\frac{s_d}{\sqrt{n}}} =\frac{-4.125-(-5)}{\frac{1.246}{\sqrt{8}}}=1.986$

Since $t-1.986 >-1.895$ , The null hypothesis fails to reject. The results do not support the packaged food producer's claim with a Type 1 error probability of 0,05.

From t table, at $\alpha= 0.01$ , reject null hypothesis if $t$ .

Since $t=1.986>-2.998$ , fail to reject null hypothesis.

Data do not support the claim of the producer of the dietary product with the probability of Type 1 error of .01.

5 0

1 year ago

During the 2015-16 NBA season, J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 . Assume th

ser-zykov [4K]

Answer: 0.5898

Step-by-step explanation:

Given : J.J. Redick of the Los Angeles Clippers had a free throw shooting percentage of 0.901 .

We assume that,

The probability that .J. Redick makes any given free throw =0.901 (1)

Free throws are independent.

So it is a binomial distribution .

Using binomial probability formula, the probability of getting success in x trials :

$P(X=x)^nC_xp^x(1-p)^{n-x}$

, where n= total trials

p= probability of getting in each trial.

Let x be binomial variable that represents the number of a=makes.

n= 14

p= 0.901 (from (1))

The probability that he makes at least 13 of them will be :-

$P(x\geq13)=P(x=13)+P(x=14)$

$=^{14}C_{13}(0.901)^{13}(1-0.901)^1+^{14}C_{14}(0.901)^{14}(1-0.901)^0\\\\=(14)(0.901)^{13}(0.099)+(1)(0.901)^{14}\ \ [\because\ ^nC_n=1\ \&\ ^nC_{n-1}=n ]\\\\\approx0.3574+0.2324=0.5898$

∴ The required probability = 0.5898

5 0

2 years ago

Consider a disease whose presence can be identified by carrying out a blood test. Let p denote the probability that a randomly s

kifflom [539]

Answer:

The expected number of tests, E(X) = 6.00

Step-by-step explanation:

Let us denote the number of tests required by X.

In the case of 5 individuals, the possible value of x are 1, if no one has the disease, and 6, if at least one person has the disease.

To find the probability that no one has the disease, we will consider the fact that the selection is independent. Thus, only one test is necessary.

Case 1: P(X=1) = [P (not infected)]⁵

= (0.15 - 0.1)⁵

P(X=1) = 3.125*10⁻⁷

Case 2: P(X=6) = 1- P(X=1)

= 1 - (1 - 0.1)⁵

P(X=6) = (1 - 3.125*10⁻⁷) = 0.999999

P(X=6) = 1.0

We can then use the previously determined values to compute the expected number of tests.

E(X) = ∑x.P(X=x)

= (1).(3.125*10⁻⁷) + 6.(1.0)

E(X) = E(X) = 6.00

Therefore, the expected number of tests, E(X) = 6.00

3 0

2 years ago

which equation is in slope-intercept form and represents a line with slope 3 through the point (9,-4)

krek1111 [17]

Slope-intercept form is y = mx +b, where the variable m represents the slope of the line and the variable b represents the y-intercept of the line. Since we weren't given the y-intercept, one way to solve this problem is substitute the slope for the variable m and then the x and y values from the ordered pair to solve for b.

y = mx + b

y = 3x + b

-4 = 3(9) + b

Now, we can simplify by computing the multiplication on the right side of the equation.

-4 = 27 + b

Finally, we can find the value for b by subtracting 27 from both sides of the equation to get:

-31 = b

Now, we should substitute this value for b back into our equation with the slope.

y = 3x - 31

Therefore, your answer is y = 3x - 31.

Hope this helps!

6 0

2 years ago

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