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3 years ago

The main goal of a personal budget is to

Mathematics

2 answers:

Sunny_sXe [5.5K]3 years ago

7 0

Answer:

have income be greater than expenses

Step-by-step explanation:

Evgesh-ka [11]3 years ago

3 0

Make sure you don't spend to much and have no more money left

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If the discriminant, b^2-4ac is ___, a quadratic will have two real roots, two points of intersection with the x-axis.

den301095 [7]

If the discriminant, b² - 4ac is<u> positive</u>

<h3>How to complete the statement?</h3>

From the question, we have the following equation of discriminant

The discriminant (d) is calculated as

d = b² - 4ac

The solutions of a quadratic equation are dependent on the following conditions

If d = 0, the quadratic equation has 1 real solution
If d < 0, the quadratic equation has imaginary solutions
If d > 0, the quadratic equation has 2 different real solutions

This means that "d > 0, the quadratic equation has 2 different real solutions" implies that the discriminant is positive

Hence, the complete statement is (b) positive

Read more about discriminant at

brainly.com/question/64103

#SPJ1

6 0

1 year ago

help please i'll give brainliest if you give me the answers to all if you scam me your getting reported

LiRa [457]

Answer:

Step-by-step explanation: I just answered for you?

8 0

2 years ago

PLEASE HELP URGENT!!!!

Setler [38]

Answer:

A hope this helps

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Can somebody prove this mathmatical induction?

Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

$\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2$

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

3 step:

Check for n=k+1 whether the statement

$\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)$

is true.

Start with the left side:

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)$

According to the 2nd step,

$\sum \limits_{j=1}^k2^j=2(2^k-1)$

Substitute it into the $\ast$

$\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)$

So, you have proved the initial statement

4 0

4 years ago

A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di

Lena [83]

Answer:

Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .

Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, $\mu$ = 1.3 cm and Standard deviation, $\sigma$ = 0.01 cm

Also, since distribution is normal;

Z = $\frac{X -\mu}{\sigma}$ ~ N(0,1)

Let X = range of diameters

So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)

P(X < 1.33) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.33 -1.3}{0.01}$ ) = P(Z < 3) = 0.99865

P(X <= 1.27) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.27 -1.3}{0.01}$ ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

= 0.00135

P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

7 0

3 years ago