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2 years ago

Consider the conditional statement shown.

2 answers:

6 0

Hello!

First of all, 6 and 10 are not prime numbers as they both have at least two pairs of factors (one factor pair plus one and itself).

Anything multiplied by two is even. Therefore, a) 2 counterexamples the statement. Just to show why 7 doesn't work, seven multiplied by 3 is 21, which is odd, and 7 times 7 is 49, and 7 times 5 is 35 (all prime multipliers). The only one that works is two.

I hope this helps!

miv72 [106K]2 years ago

6 0

Answer: 2

Step-by-step explanation:

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The second question

stich3 [128]

A: 5

B: 9a & a (from a/6)

C: -5 & ÷4

D: -5 & 9a

I haven't done algebra in a year, so don't think my answers are perfect!

definitions:

term: something separated by a sign/symbol (÷, ×, - +) (a/6 are two separate terms, ÷)

constant terms: variables that can be solved.

unlike terms: terms that don't "go" together, you can't subtract 5 from 9a because there's a variable in the way (eyy that rhymes)

like terms: terms that you can add/subtract/multiply/divide to another term

(another answer to c is 9a & a)

7 0

3 years ago

The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the

lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

Let $\mu$ = average sodium content.

So, Null Hypothesis, $H_0$ : $\mu$ = 1100 milligrams {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 1100 milligrams {means that the sodium content is different from what the nutrition label states}

The test statistics that would be used here One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n}}}$ ~ $t_n_-_1$

where, $\bar X$ = sample average sodium content = 1088.64 milligrams

s = sample standard deviation = 234.12 milligrams

n = sample of packages of Oriental Spice Sauce = 40

So, test statistics = $\frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}$ ~ $t_3_9$

= -0.307

The value of z test statistics is -0.307.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the sodium content is same as what the nutrition label states.

3 0

3 years ago

Hellooooo everyone its donald trump!! ihave a question?What led to the decline of the sugar business in the West Indies?

lbvjy [14]

D) doctors told peopel thag suger would damage their health

5 0

3 years ago

Read 2 more answers

Prove the identity (sina-cosatanb)/(cosa+sinatanb) = tan (a+b)

OLga [1]

Sina - (cosa)(tanb)/cosa + (sina)(tanb)
sina ≡ (tana)(cosa)

(tana)(cosa) - (cosa)(tanb)/cosa + (tana)(cosa)(tanb)
= cosa(tana - tanb)/cosa(1 + tanatanb)
(cosas cancel out)
= (tana - tanb)/(1 + tanatanb) ≡ tan(a-b)

7 0

3 years ago

¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para

Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

$\frac{x-2}{p}=\frac{2-y}{p}$

$\frac{x-1}{1-p}=\frac{y-2}{2}$

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

$\frac{x-2}{p}=\frac{2-y}{p}$

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

$m_{1}=-1$

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

$m_{2}=-\frac{1}{m_{1}}$