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3 years ago

Solve for n. -4n + 7 = 1 – 2n

Mathematics

1 answer:

Solnce55 [7]3 years ago

3 0

Answer:

n = 3

Step-by-step explanation:

subtract 1 from both sides to get:

-4n + 6 = -2n

then add 4n to both sides to get:

6 = 2n

then divide both sides by 2 to get:

3 = n

You might be interested in

Find the zeros of the function g(x) = 2x2 - 5x + 2.

sertanlavr [38]

G(x) = 2x² - 5x + 2 = 2x² - 4x - x + 2 = 2x · x - 2x · 2 - 1 · x - 1 · (-2)

= 2x(x - 2) -1(x - 2) = (x - 2)(2x - 1)

g(x) = 0 ⇔ (x - 2)(2x - 1) = 0 ⇔ x - 2 = 0 or 2x - 1 = 0

x = 2 or x = 0.5

8 0

3 years ago

I need help with this question, I can work out the ratio but I am just confused with the question. So do help if you can please.

nirvana33 [79]

Answer:

76%

Step-by-step explanation:

So basically 528 represents a full circle and the amount in the circle is 2:5 to that in the stalls.

So you divide 528 by 2 and then multiply it by 5 to get a full stall of 1320

you then add 1320 and 528 to get a full theater of 1848

to get 2/3 of seats in the stall you divide 1320 by 3 and then multiply it by 2 to get 880.

For 2/3 of the stall and a full circle you add 880 and 528 to get 1408.

You divide the amount of people in the theater on Friday by the total amount of people which the theater can hold so 1408 divided by 1848 and you get the percentage of 76%

8 0

3 years ago

The number of bacteria in a Petri dish increases by 18% every hour. If there were initially 200 bacteria placed in the Petri dis

In-s [12.5K]

Answer:

Step-by-step explanation:

an exponential equation has the form y=ar^x where y=final amount, a=initial amount, r=rate, and x=time, in this case

y=200(1.18)^x

5 0

3 years ago

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0

3 years ago

An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date o

skelet666 [1.2K]

Answer:

x1=c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)

x2=c(30,42,47,47,43,41,48,44,43,50,56,60)

y=c(210,110,103,103,91,76,73,70,68,53,45,31)

mod=lm(y~x1+x2)

summary(mod)

R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:

Min 1Q Median 3Q Max

-41.730 -12.174 0.791 12.374 40.093

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 415.113 82.517 5.031 0.000709 ***

x1 -6.593 4.859 -1.357 0.207913

x2 -4.504 1.071 -4.204 0.002292 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 24.45 on 9 degrees of freedom

Multiple R-squared: 0.768, Adjusted R-squared: 0.7164

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a). y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

c). y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101

residual =68-81.0101 = -13.0101

d). F=14.9

P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e). newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

fit lwr upr

1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f). #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

fit lwr upr

1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

g). No, there is not evidence this factor is significant. It should be dropped from the model.

4 0

3 years ago