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Morgarella [4.7K]
3 years ago
7

Use ten to subtract

Mathematics
1 answer:
JulsSmile [24]3 years ago
8 0
X - 10
in function...
v = f(X - 10)
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4a 3b - C<br> per a - 3;<br> b = -1;<br> 1; C = -2
Taya2010 [7]

Answer:

4×(-3)=12,3×(-1)=-3;+3,c=-2

3 0
3 years ago
Evaluate the expression below for x = 5. 5(x + 7)​
wariber [46]

Answer:

60

Step-by-step explanation:

5(x + 7)

when x is equal to 5, then

  • 5(x + 7)
  • 5(5+7)
  • 5(12)
  • 5×12
  • 60
7 0
2 years ago
A crowbar 29 in. long is pivoted 6 in. from the end. What force must be applied at the long end in order to lift a 700 Ib object
Alla [95]

A force of 182.6 lb should be applied to lift 700 lb object.

<h3>How to calculate the force?</h3>

Let the force required to lift be F

Total length = 29 in.

Distance of short end from pivot =  in.

Distance from long end = 29-6 = 23 in.

Now Equating moments

F × 23 = 700x6

23F = 4200

F = 4200/23

= 182.6 lb

A force of 182.6 lb should be applied to lift 700 lb object.

Learn more about force on:

brainly.com/question/12970081

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6 0
1 year ago
PLS HELP!!
irina [24]
$789.18 * 0.062 = $48.93
answer is <span>48.93 (first choice)</span>
8 0
3 years ago
Read 2 more answers
An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than
Sauron [17]

The test statistic for the hypothesis would be 1.413.

Given that the participants in its new diet program lose, on average, more than 13 pounds and the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds.

The objective is to text the advertisement's claim that participants in new diet program lose weight, on average, more than 13 pounds.

Hypothesis:

Null hypothesis:H₀:μ=13

Alternative hypothesis:Hₐ:μ>13

Here, μ be the mean weight loss of all participants.

n=30,

Degree of freedom n-1=30-1=29

\bar{x}=13.8 and s=3.1

To test the null hypothesis H₀, the value of test static would be calculated as follows:

\begin{aligned}t&=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\ t_{29}&=\frac{13.8- 13}{\frac{3.1}{\sqrt{30}}}\\ &=\frac{0.8}{0.566}\\ &=1.413\end

Hence, the value of the test static for the hypothesis with the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds is 1.413.

Learn more about hypothesis from here brainly.com/question/14783359

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6 0
1 year ago
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