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3 years ago

Can someone help me with this plzz

Mathematics

2 answers:

Bad White [126]3 years ago

4 0

14 times

Add up all the numbers in the table and you get 50. If every 50 spins it lands on 1 seven times. 50x2=100 so 7x2=14

topjm [15]3 years ago

3 0

Answer:

Step-by-step explanation:

We can see that the spinner is spun 50 times if we add all the "times spun" up.

7 + 8 + 7 + 8 + 11 + 9 = 50

So if we multiply this by 2, we should get the theoretical probability that 1 will be spun if the spinner was spun 100 times.

1 was spun 7 times when we spun the spinner 50 times, so we can predict that 1 will be spun 14 times if we spin the spinner 100 times.

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In a standard normal distribution 95% of the data is within +/- _______standard deviations of the mean.

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Four years ago, you started a job with an annual salary of $40,000. At the end of each year on the job, you received a raise of

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3 years ago

Jose is thinking of a number.

alexira [117]

The number is 66.

Solution:

Let the number be x.

To find the number x using given conditions.

Start with the last condition.

It is less than 7 times 8 + 23 = 7 × 8 + 23 = 79

So, the number is less than 79 which implies x < 79.

Using condition (4),

It is greater than 5 times 4 = 5 × 4 = 20

So, the number is greater than 79 which implies x > 20.

Combining condition (4) and (7), we get 20 < x < 79.

Using condition (3) and (5),

The number is a multiple of 11 and 6.

The number which is multiplied by both 11 and 6 is 66, 132, ...

But here x lies between 20 < x < 79.

So x = 66.

66 is not odd number and the sum of the digits is divisible by 2.

Therefore condition (1) and (2) also satisfied.

Hence the number is 66.

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3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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