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2 years ago

Solve pls brainliest

Mathematics

2 answers:

Tju [1.3M]2 years ago

7 0

Answer:

Step-by-step explanation:

5*6 = 30

30 - 8

That comes out to 22

lesya [120]2 years ago

6 0

Answer:

(5×6) -8 = 30 -8 = 22

the answer is 22

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Find 2 consecutive odd integers sum of who's square is 290 pls send fast

ELEN [110]

Answer:

<h3>Hence required integers are 11 and 13</h3>

Let x an odd positive integer

Then, according to question

x^2 +(x+2)^2=290

2x^2 +4x−286=0

x^2 +2x−143=0

x ^2+13x−11x−143=0

(x+13)(x−11)=0

x=11 as x is positive

Hence required integers are 11 and 13

Step-by-step explanation:

Hope it is helpful....

7 0

2 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 436.0 gram setting. It

Vilka [71]

Answer:

We conclude that the machine is under filling the bags.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 436.0 gram

Sample mean, $\bar{x}$ = 429.0 grams

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 23.0 grams

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 436.0\text{ grams}\\H_A: \mu < 436.0\text{ grams}$

We use one-tailed(left) z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{429 - 436}{\frac{23}{\sqrt{40}} } = -1.92$

Now, $z_{critical} \text{ at 0.05 level of significance } = -1.64$

Since,

$z_{stat} < z_{critical}$

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that the machine is under filling the bags.

8 0

3 years ago

Number 2 I need help on how to solve it

Hunter-Best [27]

Let's simplify the following expression:

$\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}$

We get,

$\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}$ $\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}$ $\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}$