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stiv31 [10]
3 years ago
7

What is the quotient of 5s ad 8

Mathematics
1 answer:
butalik [34]3 years ago
4 0

Answer:

0.625 s

Step-by-step explanation:

5s/8

0.625 s

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Help help failing help needed
Karo-lina-s [1.5K]
The difference would be 37

4 0
3 years ago
The shares of the U.S. automobile market held in 1990 by General Motors, Japanese manufacturers, Ford, Chrysler, and other manuf
Mariana [72]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the purchase frequencies of new-car buyers follow the distribution of the shares of the U.S. automobile market for 1990.

You have one variable of interest:

X: Brand a new-car buyer prefers, categorized: GM, Japanese, Ford, Chrysler and Other

n= 1000

Observed frequencies

GM 193

Japanese 384

Ford 170

Chrysler 90

Other 163

a) The test to use to analyze if the observed purchase frequencies follow the market distribution you have to conduct a Goodness to Fit Chi-Square test.

The conditions for this test are:

- Independent observations

In this case, we will assume that each buyer surveyed is independent of the others.

- For 3+ categories: each expected frequency (Ei) must be at least 1 and at most 20% of the Ei are allowed to be less than 5.

In our case we have a total of 5 categories, 20% of 5 is 1, only one expected frequency is allowed to have a value less than 5.

I'll check this by calculating all expected frequencies using the formula: Ei= n*Pi (Pi= theoretical proportion that corresponds to the i-category)

E(GM)= n*P(GM)= 1000*0.36= 360

E(Jap)= n*P(Jap)= 1000*0.26= 260

E(Ford)= n*P(Ford)= 1000*0.21= 210

E(Chrys)= n*P(Chrys)= 1000*0.09= 90

E(Other)= n*P(Other)= 1000*0.08= 80

Note: If all calculations are done correctly then ∑Ei=n.

This is a quick way to check if the calculations are done correctly.

As you can see all conditions for the test are met.

b) The hypotheses for this test are:

H₀: P(GM)= 0.36; P(Jap)= 0.26; P(Ford)= 0.21; P(Chrys)= 0.09; P(Other)= 0.08

H₁: At least one of the expected frequencies is different from the observed ones.

α: 0.05

X^2= sum \frac{(Oi-Ei)^2}{Ei} ~~X^2_{k-1}

k= number of categories of the variable.

This test is one-tailed right this mean you'll reject the null hypothesis to high values of X²

X^2_{k-1;1-\alpha }= X^2_{4;0.95}= 9.488

Decision rule using the critical value approach:

If X^2_{H_0} ≥ 9.488, reject the null hypothesis

If X^2_{H_0} < 9.488, don't reject the null hypothesis

X^2_{H_0}= \frac{(193-360)^2}{360} + \frac{(384-260)^2}{260} + \frac{(170-210)^2}{210} + \frac{(90-90)^2}{90} + \frac{(163-80)^2}{80} = 230.34

The value of the statistic under the null hypothesis is greater than the critical value, so the decision is to reject the null hypothesis.

Using a 5% level of significance, there is significant evidence to conclude that the current market greatly differs from the preference distribution of 1990.

5 0
3 years ago
I will give brainliest Please answer
faust18 [17]

Answer:

Step-by-step explanation:

1) 9 x 4 = 36

9 + 4 = 13

2) 9 x -4 = -36

9 + (-4) = 5

3) 4 x3 = 12

4+ 3 = 7

4) -4 x 3 = -12

-4 + 3 = -1

8 0
3 years ago
Each of 100 students in the Allen School can only take 1 CSE class each, between the four classes CSE 311, CSE 312, CSE 331, and
Naddika [18.5K]

Answer:

P(a=31,b=39,c=7,d=23) = 0.000668

Step-by-step explanation:

Sample space, n = 100

Let the number of students signed up for CSE 311 = a

Let the number of students signed up for CSE 312 = b

Let the number of students signed up for CSE 331 = c

Let the number of students signed up for CSE 332 = d

Probability of taking CSE 311, P_a = 0.3

Probability of taking CSE 312, P_b = 0.4

Probability of taking CSE 331, P_c = 0.1

Probability of taking CSE 332, P_d = 0.2

P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a}  p_b^{b}  p_c^{c}  p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} *  0.3^{31} * 0.4^{39} * 0.1^{7}  0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668

3 0
3 years ago
The recommended dose of Vancocin® for a severe staphylococcal infection
Dvinal [7]
180lb=81.6kg

\frac{40 mg}{1kg} = \frac{xmg}{81.6kg}

40 * 81.6 = x

3264 =x

3264mg

\frac{24hr}{6hr} = 4hr

3264*4

13056 mg/day
7 0
3 years ago
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