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3 years ago

Solve for x X over 4 minus negative nine equals ten point one X/4 - -9 = 10.1

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Answer:

x=4.4

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

14x+9=10.1

Step 2: Subtract 9 from both sides.

14x+9−9=10.1−9

14x=1.1

Step 3: Multiply both sides by 4.

4*(14x)=(4)*(1.1)

x=4.4

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How much money left after spending $8, $7 and $13, then received $10 and spent $15. now $12 is left. how much money did one have

Inessa [10]

$8 + $7 + $13 + 15 = $43
$x - $43 = $12
x = $55

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4 0

3 years ago

A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measurin

Ratling [72]

Answer:

The standard deviation of weight for this species of cockroaches is 4.62.

Step-by-step explanation:

Given : A different species of cockroach has weights that are approximately Normally distributed with a mean of 50 grams. After measuring the weights of many of these cockroaches, a lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

To find : What is the approximate standard deviation of weight for this species of cockroaches?

Solution :

We have given,

Mean $\mu=50$

The sample mean x=55

A lab assistant reports that 14% of the cockroaches weigh more than 55 grams.

i.e. P(X>55)=14%=0.14

The total probability needs to sum up to 1,

$P(X\leq 55)=1-P(X>55)$

$P(X\leq 55)=1-0.14$

$P(X\leq 55)=0.86$

The z-score value of 0.86 using z-score table is z=1.08.

Applying z-score formula,

$z=\frac{x-\mu}{\sigma}$

Where, $\sigma$ is standard deviation

Substitute the values,

$z=\frac{x-\mu}{\sigma}$

$1.08=\frac{55-50}{\sigma}$

$1.08=\frac{5}{\sigma}$

$\sigma=\frac{5}{1.08}$

$\sigma=4.62$

The standard deviation of weight for this species of cockroaches is 4.62.

4 0

3 years ago

Find the sum of the following infinite geometric series.

Crazy boy [7]

Answer:

Final answer is $\frac{80}{3}$ .

Step-by-step explanation:

We have been given an infinite geometric series.

Now we need to find it's sum.

common ratio $r=-\frac{1}{5}$ .

plug n=1 to get the first term

$a_n=32\left(-\frac{1}{5}\right)^{\left(n-1\right)}$

$a_1=32\left(-\frac{1}{5}\right)^{\left(1-1\right)}$

$a_1=32\left(-\frac{1}{5}\right)^{\left(0\right)}$

$a_1=32\left(1\right)$

$a_1=32$

Now plug these values into infinite sum formula

$S_{\infty}=\frac{a_1}{1-r}=\frac{32}{1-\left(-\frac{1}{5}\right)}=\frac{32}{1.2}=\frac{320}{12}=\frac{80}{3}$

Hence final answer is $\frac{80}{3}$ .

8 0

3 years ago

Students have organized

Olenka [21]

Answer:

Part A. At least 6 hours

Part B. In less than 2.5 hours Elijah will be behind Mercedes

Part C. In more than 2.5 hours Elijah will be ahead Aubrey

Step-by-step explanation:

D = distance

v =speed

t = time

Formula connecting D, v and t:

$D=v\cdot t$

Part A.

Steve's speed: $v=3.5\ mph$

Distance: at least 21 miles

Time: unknown, so

$3.5\cdot t\ge 21\\ \\35t\ge 210\ [\text{Multiplied by 10}]\\ \\t\ge \dfrac{210}{35}\\ \\t\ge \dfrac{30}{5}\\ \\t\ge 6$

It would take Steve at least 6 hours to walk at least 21 mi on Day 1.

Part B.

Mercedes's speed: $v_M=2.4\ mph$

Elijan's speed: $v_E=3.2\ mph$

Elijan's Distance walked: $D_E$ miles

Mercedes's Distance walked: $D_M$ miles

Time: x hours

Mercedes is 2 miles ahead, so

$D_E=3.2x\\ \\D_M=2.4x+2$

Elijan will be behind when

$D_E$

In 2.5 hours Elijan will catch up Mercedes, and in less than 2.5 hours Elijah will be behind Mercedes.

Part C.

Aubrey's speed: $v_M=3\ mph$

Elijan's speed: $v_E=3.2\ mph$

Elijan's Distance walked: $D_E$ miles

Aubrey's Distance walked: $D_A$ miles

Time: x hours

At the beginning of Day 3, Elijah starts walking at the marker for Mile 42, and Aubrey starts walking at the marker for Mile 42.5.

$D_E=42+3.2x\\ \\D_A=42.5+3x$

Elijan will be ahead of Aubrey when

$D_E>D_A\\ \\42+3.2x> 42.5+3x\\ \\3.2x-3x>42.5-42\\ \\0.2x>0.5\\ \\2x>5\ [\text{Multiplied by 10}]\\ \\x>\dfrac{5}{2}\\ \\x>2.5\ hours$

In 2.5 hours Elijan will catch up Aubrey, and in more than 2.5 hours Elijah will be ahead Aubrey.

4 0

3 years ago

It takes 26 pounds of seed to completely plant a 3-acre field. How many pounds of seed are needed per acre?

satela [25.4K]

Answer:

3.25 pounds per acre

Step-by-step explanation:

We want to find how much pounds of seed is need per acre. In division, with a numerator and denominator, this works perfectly, with how much we need (of seed) per acre translating into $\frac{26 pounds}{3 acres}$ , with the line representing the per. We can plug this into a calculator to get 3.25 pounds per acre, or $\frac{3.25 pounds}{1 acre}$

3 0

3 years ago