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3 years ago

I usually dont ask for help in math, but this has me stumped. pls help!

Mathematics

1 answer:

USPshnik [31]3 years ago

7 0

Use photomath it's an app used for mathematic problems but not word problems

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A golfers score after playing on friday was +2

jek_recluse [69]

The end of his round on Saturday was a -5, and the end of his round on Sunday was a 0.

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3 years ago

Pls help I don't understand:

sineoko [7]

Answer:

Part a) $T(d)=2d+30$

Part b) $T(6)=42\ minutes$

Step-by-step explanation:

Part a) Write an equation for T (d)

Let

d ----> the number of days

T ---> the time in minutes of the treadmill

we know that

The linear equation in slope intercept form is equal to

$T=md+b$

where

m is the slope or unit rate

b is the y-intercept or initial value

In this problem we have

The slope or unit rate is

$m=2\ \frac{minutes}{day}$

The y-intercept or initial value is

$b=30\ minutes$

substitute

$T(d)=2d+30$

Part b) Find T (6), the minutes he will spend on the treadmill on day 6

For d=6

substitute in the equation and solve for T

$T(6)=2(6)+30$

$T(6)=42\ minutes$

8 0

3 years ago

$35,000 at 2% for 9 years. ending balance and interest

love history [14]

Answer:

$Solution$

Here,

$Principle(p)=$35000\\Rate(r)=2%\\\\Time(t)=9 years$

$Now, Interest(I)=\frac{PTR}{100}$

$=\frac{35000*2*9}{100}\\= $3600$

6 0

2 years ago

A kite, flying 50 feet high in the air is attached by a string to a stake in the sand. How long is the string to the nearest ten

Vlad [161]

Height of the kite from the ground = 50 feet
Angle at which the kite is flying = 45 degrees
Let us assume the length of the string = x feet
Then
x = 50 * (square root 2)
= 50 * 1.414
= 70.7 feet
From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the second option or option "b".

6 0

3 years ago

Construct a 99% confidence interval for the population standard deviation of white blood cell count (in cells per microliter). A

kogti [31]

Answer:

The correct option is 1.148 < σ < 6.015

Explanation:

The 99% confidence interval for the standard deviation is given below:

$\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\frac{0.01}{2} }} }$