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3 years ago

Helppp pleaseee asap :((((((

Mathematics

1 answer:

tangare [24]3 years ago

5 0

We are given the following:

$3(x+1)+4(x+1)=49$

Simplify the left side of the equation by using the distributive property. After you distribute, you end up with:

$3x+3+4x+4=49$

Combine like terms:

$7x + 7 = 49$

Subtract 7 from both sides to cancel the +7 on the left. When you do that, you end up with:

$7x = 42$

Divide both sides by 7 to isolate x.

$\frac{7x}{7} =\frac{42}{7}$

Your final answer should be:

$x=6$

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For a particular pickup truck the percent markup is known to be 115% based on cost to the seller. If the seller paid $15,800 for

nexus9112 [7]

Answer:

115%(15,800)= $18,170. <--- markup price

$15,800+$18,170= $33,970 total price

That is 115% x $15,800 = $18,170 markup price

Then add the original price to the markup price

That is $15,800 + $18,170 = $33,970

8 0

3 years ago

The length of a rectangle is 3 more than the width fine the length if the perimeter is 98 inches

german

Answer - Width 23

Length - 26

W = Width

L + 3 = Length

Perimeter = 2L = 2W

98 = 2(W + 3) + 2W

98 = 2W + 6 + 2W

98 = 4W + 6

92 = 4W

W = 92/4

W = 23

Width is 23 inches

Length is 3 more -- 23 + 3 = 26

We can double check our work:

26 + 26 + 23 + 23 = 98 Inches

Best of luck,

The Artemis Wolf

3 0

3 years ago

George is tracking the growth of his tomato plant. It is currently 13 cm tall and grows 2 cm per day.

Hoochie [10]

Answer:

growth = 2 x day

^growth based off what day it is

^how much it grows per day

^ day at hand

2x14 (2 weeks)= 28 cm

7 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

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

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

The teacher asked 4 students to measure 4 different objects. Here are the results

posledela

Answer:

Student 3 had the greatest percentage error.

Step-by-step explanation:

Student 1

Actual length = 50in

Measured length = 49in

Error = 50 - 49 = 1

Error% = 1/50 × 100

= 1 × 2

= 2%

Student 2

Actual length = 100cm

Measured length = 110cm

Error = 100 - 110 = 10

Error% = 10/100 × 100

= 10 × 1

= 10%

Student 3

Actual length = 2cm

Measured length = 2.5cm

Error = 2 - 2.5 = 0.5

Error% = 0.5/2 × 100

= 0.5 × 50

= 25%

Student 4

Actual length = 10ft

Measured length = 9½ft

Error = 10 - 9½ = ½

Error% = ½/10 × 100

= ½ × 10

= 1 × 5

= 5%

#teamtrees#WAP

7 0

3 years ago