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3 years ago

Solve the equation. Check your answers.29=7x+5-4x

Mathematics

1 answer:

Vladimir [108]3 years ago

6 0

Answer:

x = 8

Step-by-step explanation:

29=7x + 5 -4x

add like terms: 29 = 3x +5

get your x variable on one side

24 = 3x

divide by 3

x = 8

check!

29 = 7(8) + 5 -4(8)

29 = 56 + 5 -32

29 = 51 - 32

29 = 29

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4 years ago

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3 years ago

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has

Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e. $P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e. $P(\bar{X_2}|Y)=0.7$

So, $P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )= $1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )= $1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b) $P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)= $P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$

3 0

3 years ago

The weights of soy patties sold by a diner are normally distributed. A random sample of 15 patties yields a mean weight of 3.8 o

MariettaO [177]

Answer:

Step-by-step explanation:

Mean weigth is 3.8

standard deviation 15 =0.5

sd2 =15*(sd₁₅ )²

sd=(15*0.25)^0.5

sd=1.94

z=(3.8-4)/((0.5)/15^0.5)

z=-1.55

p=0.060571

0.060571>.05

true mean weight is less than 4 ounces

test statistic= (sample mean-mean)/((sd of sample)/n^0.5)

test statistic=(3.8-4)/((0.5)/15^0.5)

test statistic=-1.55

6 0

3 years ago

You are mixing two kinds of candy to make 10 pounds of a mixture worth $5.50 per pound. One kind is $6 per pound and the other i

Feliz [49]

Answer:

5.60 pounds each

Step-by-step explanation:

This is the dry mixture problem.

Let x = the number of pounds of the first type of candy

Therefore, number of pounds of second type of candy = 10-x

Value of first candy + value of second candy = value of mixture

Value of any candy = cost per pound of candy * weight of candy

Thus:

4x + 6(10 - x) = 5.60(10)

4x + 60 - 6x = 56

-2x + 60 = 56

-2x = -4

x = -4/-2

x = 2

Therefore, the number of pounds of first type of candy = 2 pounds

The number of pounds of second type of candy = 10 - 2 = 8 pounds

Check

$4(2) + $6(8) =

8 + 48 = $5.60(10)

$56 = $56

6 0

3 years ago

Solve the equation. Check your answers.29=7x+5-4x​

Solve the equation. Check your answers.29=7x+5-4x