1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Scrat [10]
4 years ago
7

What is the probability of getting a 7 of spades in a deck of 52 cards?

Mathematics
1 answer:
larisa86 [58]4 years ago
8 0

Answer:

I'm sure it's 1/52

Step-by-step explanation:

because if there is 52 cards and one 7 of spades then it would be a 1 out of 52 chance

You might be interested in
Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot
SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

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

The required probability = P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473

Hence, the required probability = 0.0473

5 0
3 years ago
Someone please help! im kinda in a rush
ASHA 777 [7]

Answer:

0.8 and 0.9

Step-by-step explanation:

rounded to the tenths, its 0.7

so 0.8 and 0.9 are the smallest and largest number

5 0
3 years ago
Consider the hypothesis test below.
Pavel [41]

Answer:

Z_{H0} =1.79

p-value: 0.0367

Decision: Reject H₀

Step-by-step explanation:

Hello!

Hypothesis to test:

H₀:ρ₁-ρ₂=0

H₁:ρ₁-ρ₂>0

The statistic to use to test the difference between two population proportions is the approximation of Z

Z=<u>     (^ρ₁-^ρ₂)-(ρ₁-ρ₂)     </u>   ≈N(0;1)

   √ (<u>^ρ₁(1-^ρ₁))/n₁)+(^ρ₂(1-^ρ₂)/n₂))</u>

                   

Z=<u>                    (0.28-0.15)-0                    </u>= 1.79

   √ (<u>0.28(1-0.28)/200)+(0.15(1-0.15)/300)</u>

p-value

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

P(Z>1.79)= 0.0367

Conclusion:

Comparing the p-value against the significance level, you can decide to reject the null hypothesis.

I hope you have a SUPER day!

8 0
3 years ago
Help<br> help please please
Ahat [919]

Answer:

1. A

2. C

Step-by-step explanation:

I could be wrong, that's my best guess. Im so sorry if im wrong, have a good day

7 0
3 years ago
90x+26+X<br> What is the answer
liberstina [14]

Answer:

Step-by-step explanation:

Well you can make the put the x with 90x and that will make 91x. Then you can add 91x with 26. 91x+26= 13(7x + 2). Hope that helped

7 0
3 years ago
Read 2 more answers
Other questions:
  • Algebra 1 A Semester Exam<br> 30. Write y=-3/4x-6 in standard form using integers. (1 point)
    9·1 answer
  • What is 8,147 ÷ 24 with a remainder
    8·1 answer
  • Which are perfect squares? Check all that apply.
    11·1 answer
  • Suppose a test of H0: μ = 0 vs. Ha: μ ≠ 0 is run with α = 0.05 and the P-value of the test is 0.052. Using the same data, a conf
    8·1 answer
  • A. What is an equation of the line with slope -1/2 and y-intercept 2??
    5·1 answer
  • What is 62 less than the product 22 and 3?​
    8·1 answer
  • PLZ HELP WILL GIVE BRAINLIEST
    7·1 answer
  • Please help!!!!! Its timed !!!!! Ill give out most brainly and extra points !!!
    7·1 answer
  • Complete each congruence statement by naming the corresponding angle or side (odd questions only)
    11·1 answer
  • _____=amount received after redemption -amount invested
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!